MATHEMATICS
HIGHER SECONDARY  FIRST YEAR
VOLUME  I
REVISED BASED ON THE RECOMMENDATIONS OF THE
TEXT BOOK DEVELOPMENT COMMITTEE
Untouch ability is a sin
Untouchability is a crime
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'■">* THOPoW"
TAMILNADU
TEXTBOOK CORPORATION
COLLEGE ROAD, CHENNAI  600 006
PREFACE
This book is designed in accordance with the new guidelines and
syllabi  2003 of the Higher Secondary Mathematics  First Year,
Government of Tamilnadu. In the era of knowledge explosion, writing a
text book on Mathematics is challenging and promising. Mathematics
being one of the most important subjects which not only decides the
career of many young students but also enhances their ability of
analytical and rational thinking and forms a base for Science and
Technology.
This book would be of considerable value to the students who
would need some additional practice in the concepts taught in the class
and the students who aspire for some extra challenge as well.
Each chapter opens with an introduction, various definitions,
theorems and results. These in turn are followed by solved examples
and exercises which have been classified in various types for quick and
effective revision. The most important feature of this book is the
inclusion of a new chapter namely 'Functions and Graphs'. In this
chapter many of the abstract concepts have been clearly explained
through concrete examples and diagrams.
It is hoped that this book will be an acceptable companion to the
teacher and the taught. This book contains more than 500 examples
and 1000 exercise problems. It is quite difficult to expect the teacher to
do everything. The students are advised to learn by themselves the
remaining problems left by the teacher. Since the 'Plus 1' level is
considered as the foundation course for higher mathematics, the
students must give more attention to each and every result mentioned in
this book.
The chief features of this book are
(i) The subject matter has been presented in a simple and lucid
manner so that the students themselves are able to
understand the solutions to the solved examples.
(ii) Special efforts have been made to give the proof of some
standard theorems.
(iii) The working rules have been given so that the students
themselves try the solution to the problems given in the
exercise.
(iv) Sketches of the curves have been drawn wherever
necessary, facilitating the learner for better understanding of
concepts.
(v) The problems have been carefully selected and well graded.
The list of reference books provided at the end of this book will be
of much helpful for further enrichment of various concepts introduced.
We welcome suggestions and constructive criticisms from learned
teachers and dear students as there is always hope for further
improvement.
K. SRINIVASAN
Chairperson
Writing Team
SYLLABUS
(1) MATRICES AND DETERMINANTS : Matrix Algebra  Definitions, types,
operations, algebraic properties. Determinants  Definitions, properties,
evaluation, factor method, product of determinants, cofactor
determinants. (18 periods)
(2) VECTOR ALGEBRA : Definitions, types, addition, subtraction, scalar
multiplication, properties, position vector, resolution of a vector in two and
three dimensions, direction cosines and direction ratios. (15 periods)
(3) ALGEBRA : Partial Fractions  Definitions, linear factors, none of which
is repeated, some of which are repeated, quadratic factors (none of
which is repeated). Permutations  Principles of counting, concept,
permutation of objects not all distinct, permutation when objects can
repeat, circular permutations. Combinations, Mathematical induction,
Binomial theorem for positive integral /nctexfinding middle and
particular terms. (25 periods)
(4) SEQUENCE AND SERIES : Definitions, special types of sequences and
series, harmonic progression, arithmetic mean, geometric mean,
harmonic mean. Binomial theorem for rational number other than
positive integer, Binomial series, approximation, summation of Binomial
series, Exponential series, Logarithmic series (simple problems)
(15 periods)
(5) ANALYTICAL GEOMETRY : Locus, straight lines  normal form,
parametric form, general form, perpendicular distance from a point,
family of straight lines, angle between two straight lines, pair of
straight lines. Circle  general equation, parametric form, tangent
equation, length of the tangent, condition for tangent. Equation of chord
of contact of tangents from a point, family of circles  concetric circles,
orthogonal circles. (23 periods)
(6) TRIGONOMETRY : Trigonometrical ratios and identities, signs of
Tratios, compound angles A + B, multiple angles 2A, 3A, sub multiple
(half) angle A/2, transformation of a product into a sum or difference,
conditional identities, trigonometrical equations, properties of
triangles, solution of triangles (SSS, SAA and SAS types only),
inverse trigonometrical functions. (25 periods)
(7) FUNCTIONS AND GRAPHS : Constants, variables, intervals,
neighbourhood of a point, Cartesian product, relation. Function  graph
of a function, vertical line test. Types of functions  Onto, onetoone,
identity, inverse, composition of functions, sum, difference product,
quotient of two functions, constant function, linear function, polynomial
function, rational function, exponential function, reciprocal function,
absolute value function, greatest integer function, least integer function,
signum function, odd and even functions, trigonometrical functions,
quadratic functions. Quadratic inequation  Domain and range.
(15 periods)
(8) DIFFERENTIAL CALCULUS : Limit of a function  Concept, fundamental
results, important limits, Continuity of a function  at a point, in an
interval, discontinuous function. Concept of Differentiation 
derivatives, slope, relation between continuity and differentiation.
Differentiation techniques  first principle, standard formulae, product
rule, quotient rule, chain rule, inverse functions, method of substitution,
parametric functions, implicit function, third order derivatives.
(30 periods)
(9) INTEGRAL CALCULUS : Concept, integral as antiderivative, integration of
linear functions, properties of integrals. Methods of integration 
decomposition method, substitution method, integration by parts.
Definite integrals  integration as summation, simple problems.
(32 periods)
(10) PROBABILITY : Classical definitions, axioms, basic theorems, conditional
probability, total probability of an event, Baye's theorem (statement only),
simple problems. (12 periods)
CONTENTS
Page No.
Preface
Syllabus
1. Matrices and Determinants 1
1.1 Matrix Algebra 1
1.2 Determinants 14
2. Vector Algebra 39
2.1 Introduction 39
41
42
47
58
62
69
69
74
89
98
105
4. Sequence and Series 114
4.1 Introduction 114
4.2 Sequence 115
4.3 Series 117
4.4 Some Special Types of Sequences and their Series 119
4.5 Means of Progressions 121
4.6 Some special types of Series 126
2.2
Types of Vectors
2.3
Operations on Vectors
2.4
Position Vector
2.5
Resolution of a Vector
2.6
Direction Cosines and Direction Ratios
3. Algebra
3.1
Partial Fractions
3.2
Permutations
3.3
Combinations
3.4
Mathematical Induction
3.5
Binomial Theorem
5. Analytical Geometry 133
5.1 Locus 133
135
143
154
160
167
174
179
179
182
197
215
224
232
235
Objective type Questions 243
Answers 255
5.2
Straight Lines
5.3
Family of Straight Lines
5.4
Pair of Straight Lines
5.5
Circle
5.6
Tangent
5.7
Family of Circles
6. Trigonometry
6.1
Introduction
6.2
Trigonometrical Ratios and Identities
6.3
Compound Angles
6.4
Trigonometrical Equations
6.5
Properties of Triangles
6.6
Solutions of Triangles
6.7
Inverse Trigonometrical functions
1. MATRICES AND DETERMINANTS
1.1 Matrix Algebra
1.1.1 Introduction
The term 'matrix' was first introduced by Sylvester in 1850. He defined a
matrix to be an arrangement of terms. In 1858 Cay ley outlined a matrix algebra
defining addition, multiplication, scalar multiplication and inverses. Knowledge
of matrix is very useful and important as it has a wider application in almost
every field of Mathematics. Economists are using matrices for social
accounting, input  output tables and in the study of inter industry economics.
Matrices are also used in the study of communication theory, network analysis
in electrical engineering.
For example let us consider the marks scored by a student in different
subjects and in different terminal examinations. They are exhibited in a tabular
form as given below.
Tamil
English
Maths
Science
Social Science
Test 1
70
81
88
83
64
Test 2
68
76
93
81
70
Test 3
80
86
100
98
78
The above statement of marks can also be rerecorded as follows
First row
Second row
70 81 88 83 64
68 76 93 81 70
Third row L 80 86 100 98 78
First second Third Fourth Fifth
Column Column Column Column Column
This representation gives the following informations,
(i) The elements along the first, second, and third rows represent the test
marks of the different subjects,
(ii) The elements along the first, second, third, fourth and fifth columns
represent the subject marks in the different tests.
The purpose of matrices is to provide a kind of mathematical shorthand to
help the study of problems represented by the entries. The matrices may
represent transformations of coordinate spaces or systems of simultaneous
linear equations.
1.1.2 Definitions:
A matrix is a rectangular array or arrangement of entries or elements
displayed in rows and columns put within a square bracket or parenthesis. The
entries or elements may be any kind of numbers (real or complex), polynomials
or other expressions. Matrices are denoted by the capital letters like A, B, C. . .
Here are some examples of Matrices.
" 1 4 "
First Row
1
4 2
First row (Rj)
2 5
Second Row " =
6
9 4
Second row (R2)
_ 3 6 _
Third Row
3
2 6 _
Third row (R 3 )
First Second
First
Second Third
Column Column
Column
Column Column
Ci
c 2 c 3
Note : In a matrix, rows are counted from top to bottom and the columns are
counted from left to right.
i.e. (i) The horizontal arrangements are known as rows,
(ii) The vertical arrangements are known as columns.
To identify an entry or an element of a matrix two suffixes are used. The
first suffix denotes the row and the second suffix denotes the column in which
the element occurs.
From the above example the elements of A are an = 1, ayi = 4, (221 = 2,
(222 = 5, 031 = 3 and C132 = 6
Order or size of a matrix
The order or size of a matrix is the number of rows and the number of
columns that are present in a matrix.
In the above examples order of A is 3 x 2, (to be read as 3by2) and order
of B is 3 x 3, (to be read as 3by3).
In general a matrix A of order mxn can be represented as follows :
d\\ a 12 ••• a \j ■■■ a \n
A =
an an.
— a m \ a m 2
a m j
■lh
J
column
— > i row
This can be symbolically written as A = [ay] m x n
The element a ( w belongs to i row and the/ column, i being the row index
and j being the column index. The above matrix A is an m x n or mbyn matrix.
The expression m x n is the order or size or dimension of the matrix.
Example 1.1: Construct a 3 x 2 matrix whose entries are given by ay = i  2/
Solution: The general 3x2 matrix is of the form
'an a [2 ~
A = [ay] = a 2l «22 where i = 1, 2, 3 (rows), 7 = 1,2 (columns)
_«31 a 32_
It is given that a„ = i  2/
an = 1 — 2 = — 1 ai2=l4 = 3 1
a2l =22 = a22 = 2  4 =  2
a31 = 32=1 (232 = 3  4 =  1
1.1.3 Types of matrices
(1) Row matrix: A matrix having only one row is called a row matrix or a row
vector.
Examples (i) A = [a y ]i x3=[l 7 4] is a row matrix of order 1x3.
(ii) B = [by]i x 2 = [5 8] is a row matrix of order 1x2
(iii) C = [cy]i x 1 = [100] is a row matrix of order lxl
(2) Column matrix:
A matrix having only one column is called a column matrix or a column
vector.
.". The required matrix is A =
Examples (i) A = [a,y]3 x 1 =
(ii) B = [& y ] 2 xi =
' 1
7
.4.
'25"
.30.
is a column matrix of order 3x1
is a column matrix of order 2x1
(iii) C = [Cy]i x 1 = [68] is a column matrix of order lxl
Note : Any matrix of order 1 x 1 can be treated as either a row matrix or a
column matrix.
(3) Square matrix
A square matrix is a matrix in which the number of rows and the number of
columns are equal. A matrix of order «x«is also known as a square matrix of
order n.
In a square matrix A of order n x n, the elements a\i, ai% (233 ... a nn are
called principal diagonal or leading diagonal or main diagonal elements.
"2 41
is a square matrix of order 2
A = [aij] 2 x 2
B = [bijh x 3 =
6 8.
1 2 3'
4 5 6
7 8 9.
is a square matrix of order 3.
Note: In general the number of elements in a square matrix of order n is n . We
can easily verify this statement from the above two examples.
(4) Diagonal Matrix:
A square matrix A = [ay] n x n is said to be a diagonal matrix if au = when
In a diagonal matrix all the entries except the entries along the main
diagonal are zero.
For example A = [a^ x 3 =
"4
0"
5
_0
6_
is a diagonal matrix.
(5) Triangular matrix: A square matrix in which all the entries above the
main diagonal are zero is called a lower triangular matrix. If all the entries
below the main diagonal are zero, it is called an upper triangular matrix.
A =
"3
2
7"
5
3
_0
1.
is an upper triangular matrix and B =
2
0"
4
1
8
5 7_
is a lower
triangular matrix.
(6) Scalar matrix:
A square matrix A = [ay] n x n 1S sa id to be scalar matrix if
[a if i=j
a 'J[0 if i*j
i.e. A scalar matrix is a diagonal matrix in which all the entries along the
main diagonal are equal.
B = [bij] 3 x 3
A = [aijh x 2 =
5 0'
.0 5.
■45
Vs.
are examples
for scalar matrices.
(7) Identity matrix or unit matrix:
A square matrix A = [a ( w] n x « is said to be an identity matrix if
_ri if t=j
a 'J~{0 if i*j
i.e. An identity matrix or a unit matrix is a scalar matrix in which entries
along the main diagonal are equal to 1. We represent the identity matrix of
order n as I„
h =
"1 0"
"1 0"
_0 1_
. 13 =
1
_o 1.
are identity matrices.
"0 0"
"0 0"
5
_0 0_
_0 0_
are examples of zero matrices.
(8) Zero matrix or null matrix or void matrix
A matrix A = [a,y] m x « is said to be a zero matrix or null matrix if all the
entries are zero, and is denoted by O i.e. ay = for all the values of i,j
"0 0"
[0 0],
_0 0_
(9) Equality of Matrices:
Two matrices A and B are said to be equal if
(i) both the matrices A and B are of the same order or size.
(ii) the corresponding entries in both the matrices A and B are equal.
i.e. the matrices A = [a,)] m x n an d B = [bjj] p x q are equal if m = p, n = q
and ajj = by for every i andj.
Example 1.2 :
[x y~\ [4 31
If = ,  then find the values of x, y, z, w.
\_z w_l LI 5J J
Solution:
Since the two matrices are equal, their corresponding entries are also equal.
:.x = A y = 3 z=l w = 5
(10) Transpose of a matrix:
The matrix obtained from the given matrix A by interchanging its rows
into columns and its columns into rows is called the transpose of A and it is
T
denoted by A' or A .
"4 3"
If A =
then A T =
2 1"
5.
,T
Note that if A is of order m x n then A is order nxm.
(11) Multiplication of a matrix by a scalar
Let A be any matrix. Let k be any nonzero scalar. The matrix kA is
obtained by multiplying all the entries of matrix A by the non zero scalar k.
i.e. A = yo.ij\ m x n — ^ kA. = yka\j\ m x n
This is called scalar multiplication of a matrix.
Note: If a matrix A is of order m x n then the matrix kA is also of the same
order m x n
For example If A =
(12) Negative of a matrix:
Let A be any matrix. The negative of a matrix A is  A and is obtained by
changing the sign of all the entries of matrix A.
16 A = i&ij\ m x n —? ~ A = \ — @ij\m x n
1 7 2"
then 2A = 2
"17 2"
" 2 14 4
■6 3 9.
_6 3 9_
_ 12 6 18
Let A =
then
■A =
cos9
sin9
 sin9
cos9.
cos9 sin9
 sin9 cos9_
1.1.4 Operations on matrices
(1) Addition and subtraction
Two matrices A and B can be added provided both the matrices are of the
same order and their sum A + B is obtained by adding the corresponding entries
of both the matrices A and B
i.e. A = [aij] m x „ and B = [bij] m x „ then A + B = [ay + bij\ m x „
Similarly A  B = A + ( B) = [a,y] w x „ + [ by] m x „
= \flij ~ Dij\m x n
Note:
(1) The matrices A + B and A  B have same order equal to the order of
AorB.
(2) Subtraction is treated as negative addition.
(3) The additive inverse of matrix A is  A.
i.e. A + ( A) = ( A) + A = O = zero matrix
For example, if A =
"7
2 "
4
7"
8
6
andB =
3
1
_9
6.
_8
5_
then A + B =
AB = A + (B) =
"7 2 "
4
7"
8 6
+
3
1
=
_9 6.
_8
5_
7 + 4
8 + 3
98
27"
"11
5"
6+1
=
11
7
6 + 5_
1
1.
~7
2 "
"4
7"
8
6
+
3
1
=
_9
6.
_ 8
5.
74
83
9 + 8
and
2 + 7"
" 3
9"
61
=
5
5
65_
.17
11.
(2) Matrix multiplication:
Two matrices A and B are said to be conformable for multiplication if the
number of columns of the first matrix A is equal to the number of rows of the
second matrix B. The product matrix 'AB' is acquired by multiplying every row
of matrix A with the corresponding elements of every column of matrix B
elementwise and add the results. This procedure is known as rowbycolumn
multiplication rule.
Let A be a matrix of order m x n and B be a matrix of order n x p then the
product matrix AB will be of order m x p
i.e. order of A is m x n, order of B is n x p
Aiumber of rows^ /number of columns^
Then the order of AB is m x p = [ rf matrk A )x{ of matrix B )
The following example describes the method of obtaining the product
matrix AB
Let A =
'2 1 4'
7 3 6.
B =
2x3
6 4 3'
3 2 5
.7 3 1.
3x3
It is to be noted that the number of columns of the first matrix A is equal to
the number of rows of the second matrix B.
.". Matrices A and B are conformable, i.e. the product matrix AB can be
found.
"6 4 3"
"2 1 4"
AB =
1 3 6_
3 2 5
_7 3 1.
2 1 4
7 3 6
2 1 4
2 1 4
7 3 6
6 7 3 6 4
3 2
7 3
"(2) (6) + (1) (3) + (4) (7) (2) (4) + (1) (2) + (4) (3)
L(7) (6) + (3) (3) + (6) (7) (7) (4) + (3) (2) + (6) (3)
"12 + 3 + 28 8 + 2+12 6 + 5 + 4 "1
_42 + 9 + 42 28 + 6+18 21 + 15 + 6.
It is to be noticed that order of AB is 2 x 3, which is the number of rows of
first matrix A 'by' the number of columns of the second matrix B.
Note : (i) If AB = AC, it is not necessarily true that B = C. (i.e.) the equal
matrices in the identity cannot be cancelled as in algebra,
(ii) AB = O does not necessarily imply A = O or B = O
3
5
1 I
(2) (3) + (1) (5) + (4) (1)
(7) (3) + (3) (5) + (6) (1)J
"43 22 15
_93 52 42.
AB:
For example, A =
but AB =
1 1
1 1
"1 1
1 1
* O and B =
1 1"
1 1.
*o
0'
.0 0.
= o
(iii) If A is a square matrix then A. A is also a square matrix of the
same order. AA is denoted by A . Similarly A A = AAA = A
If I is a unit matrix, then I = I 2 = I 3 = . . . = I".
1.1.5 Algebraic properties of matrices:
(1) Matrix addition is commutative:
If A and B are any two matrices of the same order then A + B = B + A.
This property is known as commutative property of matrix addition.
(2) Matrix addition is associative:
i.e. If A, B and C are any three matrices of the same order
thenA+(B + C) = (A+B)+C. This property is known as associative property
of matrix addition.
(3) Additive identity:
Let A be any matrix then A + = + A = A. This property is known as
identity property of matrix addition.
The zero matrix O is known as the identity element with respect to matrix
addition.
(4) Additive inverse:
Let A be any matrix then A + ( A) = ( A) + A = O. This property is
known as inverse property with respect to matrix addition.
The negative of matrix A i.e.  A is the inverse of A with respect to matrix
addition.
(5) In general, matrix multiplication is not commutative i.e. AB * BA
(6) Matrix multiplication is associative i.e. A(BC) = (AB)C
(7) Matrix multiplication is distributive over addition
i.e. (i) A(B + C) = AB + AC (ii) (A + B)C = AC + BC
(8) AI = IA = A where I is the unit matrix or identity matrix. This is known as
identity property of matrix multiplication.
Example 1.3: If A =
Prove that
"1 8"
"1 3"
~4
6"
_4 3_
B =
1 4_
C =
. 3
5_
(i) AB * BA
(iii) A(B + C) = AB + AC
(ii) A(BC) = (AB)C
(iv) AI = IA = A
Solution:
(i)
AB =
BA =
1 8
4 3
'1 + 56
.4 + 21
1 3
.7 4
1 + 12
7 + 16
T 3'
7 4.
3 + 32"
12 + 12.
T 8'
4 3.
8 + 9 "
56 + 12.
(1) (1) + (8) (7)
.(4) (1) + (3) (7)
'57 35"
25 24_
(1) (1) + (3) (4)
.(7) (1) + (4) (4)
13 17"
.23 68_
(1) (3) + (8) (4)'
(4) (3) + (3) (4).
...(1)
(1) (8) + (3) (3)'
(7) (8) + (4) (3).
• ••(2)
From (1) and (2) we have AB * BA
(ii) (AB)C =
"57
25
35"
24.
4
3
from (1)
(57) ( 4) + (35) (3)
.(25) (4) + (24) (3)
(57) (6) + (35) (5)
(25) (6) + (24) (5).
(AB)C =
BC =
BC =
A(BC) =
228
+ 105 342  175
_  100 + 72
150  120.
 123 167"
 28 30 _
...(3)
1 3"
"4 6"
7 4_
_ 3 5_
(l)(4) + (3)(3)
(1) (6) + (3) ( 5)"
4 + 9 615
(7) (4) + (4) (3)
(7) (6) + (4) ( 5).
_ 28 +12 42  20
59"
16
22_
1 8'
4 3.
5
16
9
22.
(1) (5)
(4) (5)
"123
_ 28
From (3) and (4) we have, (AB)C = A(BC)
A(BC) =
+ (8) (16) (l)(9) + (8)(22)
+ (3) (16) (4) (9) + (3) (22).
167'
30.
5128
2048
■9+176
36 + 66.
•••(4)
(iii) B + C =
A(B + C) =
A(B + C) =
AB =
AC =
AB + AC =
"1 3"
_7 4_
"1 8"
_4 3_
H
h
"4 6
_ 3 5
3 9"
10 1.
14 3 + 6"
"3
9"
7 + 3 45_
_ 10
1.
3 + 80 98"
12 + 30 36
3.
"77
1"
.18
33.
"57
35"
.25
24_
(5)
"1
_4
"57
_25
"77
.18
35"
24_
1"
33.
... from (1)
4 6
35.
20
7
34
9.
24 6  40"
':
20
34"
f 9 24  15.
i
9_
57 + 20 35  34"
257 2
i +
9_
(6)
10
From equations (5) and (6) we have A(B
+ C) =
= AB +
AC
(iv) Since order of A is 2 x 2, take I =
"1 0"
_0 1.
AI =
"1 8"
_4 3_
"1 0"
_0 1.
=
"1(1) + 8(0)
4(1) + 3(0)
1(0) + 8(1)"
4(0) + 3(1).
=
=
"1 8"
_4 3_
= A
IA =
"1 0"
1
"1 8"
4 3
=
"1(1) + 0(4)
0(1) + 1(4)
1(8) + 0(3)"
0(8) + 1(3).
=
=
"1 8"
_4 3_
= A
.". From(7)
and (8
) i
VI =
IA = A
'1+0 + 8"
4 + + 3_
...(7)
1+0 8 + 0"
+ 4 + 3_
...(8)
Example 1.4: If A =
'2 3'
4 5.
find A  7A  21
Solution:
A = AA =
A 2 =
'2 3'
4 5.
'2 3'
4 5.
'4 + 12 6 + 15'
.8 + 20 12 + 25.
7A =
21 =
"16
21"
_28 37_
7
"2 3'
_4 5.
=
'14
_28
21"
35_
2
"1 0"
_0 1.
=
"2
_
0"
2_
(1)
(2)
(3)
(1) + (2) + (3) gives A 2  7 A  21 = A 2 + ( 7 A) + ( 21)
16 21"
.28 37.
14 21
28 35.
2
i.e.
■ 7A  21 =
16142 2121+0
.2828 + 37352.
0'
.0 0.
= o
Example 1.5: If A =
1 4'
.0 3.
andB =
'5 0'
.3 9.
show that (A + B) 2 * A 2 + 2AB + B 2
11
Solution: A + B =
1 4'
.0 3.
'5 0'
.3 9.
1 + 5 4 + 0"
.0 + 3 3 + 9.
(A + B) z = (A + B) (A + B) =
(A + B) 2 =
,2
6 4"
.3 12.
6 4'
.3 12.
6 4'
3 12.
36 +12 24 + 48'
18 + 36 12 + 144.
'48 72"
.54 156.
(1)
A" = A.A =
B 2 = B.B =
AB =
1 4'
.0 3.
1 4'
3.
5 0"
3 9_
"5
.3 9
1 4
.0 3.
5 0"
3 9.
2AB =2
A 2 + 2AB + B 2 =
A 2 + 2AB + B 2 =
5+12 + 36
+ 9 + 27
34 72'
18 54.
25 0"
42 81.
1+0 4+12
0+0 0+9
"25 + +
.15 + 27 + 81
17 36
"1
16"
_0 9_
0"
"25 0"
Sl_
=
_42
81.
17 36
9 27.
1 161 [34 72
9_I + Ll8 54
60 88
60 144
From (1) and (2) we have
(A + B) 2 * A 2 + 2AB + B 2
Example 1.6: Find the value of x if [2x 3]
9 27.
1 + 34 + 25 16 + 72 + 0"
.0+18 + 42 9 + 54 + 81.
(2)
1 2'
3 0.
=
Solution: [2jc9 4jc + 0]
= O (Multiplying on first two matrices)
=> [(2x  9)x + 4x(3)] = O
=^> [2jc 2 + 3x] = O
i.e. 2jc 2 + 3x = => x(2x + 3) =
3
[2x A 9x+ \2x\ = O
Hence we have x = 0, x = ~j~
Example 1. 7: Solve: X + 2Y =
4 6'
8 10.
XY =
1 0'
■22.
12
Solution: Given X + 2Y =
" 4
6"
_8
10.
" 1
0"
_2
2_
(1)  (2)
XY =
(X + 2Y)  (X  Y) =
3Y =
Y =
...(1)
...(2)
" 4 6"
" 1 0"
_8 10.
2 2
" 3 6"
_6 12.
=*
■Yi
3 6"
6 12.
1 2'
2 4.
Substituting matrix Y in equation (1) we have
X + 2
X +
1 21 [4 6"
■2 4j = L8 10.
2 41 [ 4 6"
■4 8j = L8 10.
4 61 2 41 '2 2'
■8 lOJ ~L4 8j = L4 2.
2 2"
•4 2_
EXERCISE 1.1
(1) Construct a 3 x 3 matrix whose elements are (i) o« =i+j (ii) fly = i x 7
X =
.. x =
andY =
1 2'
•2 4.
(2) Find the values of x, y, z if
(3) If
(4) If A = , , . B =
7
.3 2a.
x 3x — y
2x + z 3y w_
2x 3x  y
2x + z 3y  w_
2 1"
4 2.
following
(i)  2A + (B + C) (ii) A  (3B  C) (iii) A + (B + C) (iv) (A + B) + C
(v)A + B (vi)B + A (vii)AB (viii) BA
3 2"
4 7_
ind x, y, z, w
"4 
.1
2
4_
and C =
'2
_ 1
3
2
find each of the
13
" 1
2
3"
"2
1
"1
1
1"
(5)
Given A =
13 4
B =
2 1 2
and C =
2 1
2
2 1.
.1 11.
_1 1
1.
verify the following results:
(i) AB * BA (ii) (AB) C = A(BC) (iii) A(B + C) = AB +
AC
'2 13"
"47 0"
(6)
Solve : 2X + Y H
 57 3
4 5 4_
= ; XY =
1 2 6
_2 8 5.
(7)
If A=
35"
4 2_
9
, show that A  5 A  14 I = O where I is the unit l
natri>
of order 2.
(8)
If A =
"3 2"
_4 2.
"1 2 2"
find k so that A 2 = kA 21
(9)
If A =
2 1 2
_2 2 1_
, show that A 2  4A  51 =
(10)
Solve for x if
2 r
c 1
2 3_
+
2x 3"
1 4_
=
"3 4"
_3 7_
"112"
X
(11)
Solve for x if [x 2 1]
1 4 1
_ 1  1  2_
2
.1.
= [0]
(12)
If A =
"1 2"
2 0_
B
"3
~_1
1"
0.
verify
th
e folio
win
g:
(i) (A + B) 2 = A 2 + AB + BA + B 2 (ii) (A  B) 2 * A 2  2AB + B 2
(iii) (A + B) 2 * A 2 + 2AB + B 2
(iv) (A  B) 2 = A 2  AB  BA + B 2
(v) A 2  B 2 * (A + B) (A  B)
(13) Find matrix C if A =
'3 7'
2 5.
B =
■3 2
41.
and5C + 2B =A
(14) IfA =
1 1
2 1.
andB =
and (A + B) 2 = A 2 + B 2 find x and y.
14
1.2 Determinants
1.2.1 Introduction:
The term determinant was first introduced by Gauss in 1801 while
discussing quadratic forms. He used the term because the determinant
determines the properties of the quadratic forms. We know that the area of a
triangle with vertices (jci, yi) (X2, yi) an d C*3, yj) is
2 [xi(y2y3) + x2(y3yi) + x3(yiy2)] ■•■(!)
Similarly the condition for a second degree equation in x and y to represent
2 2 2
a pair of straight lines is abc + 2fgh  af  bg  ch = . . . (2)
To minimize the difficulty in remembering these type of expressions,
Mathematicians developed the idea of representing the expression in
determinant form.
The above expression (1) can be represented in the form
Similarly the second expression (2) can be expressed as
Again if we eliminate x, y, z from the three equations
a\x + b[y + c\ z = ; ci2X +b2y + C2Z = ; aye + b^y +cj,z = 0,
we obtain ai(£>2 c 3 ~ ^3 c 2) ~ b\ (a.2 c 3 ~ a 3 c 2) + c l ( fl 2 ^3 _ a 3 i>2) =
a\ b\ c\
x\ y\ 1
form 2
X2 yi i
X3 y3 i
a h g
h b f
= 0.
8 f c
= 0. Thus a determinant is a particular
This can be written as a 2 ^2 c 2
«3 b3 c 3
type of expression written in a special concise form. Note that the quantities are
arranged in the form of a square between two vertical lines. This arrangement is
called a determinant.
Difference between a matrix and a determinant
(i) A matrix cannot be reduced to a number. That means a matrix is a
structure alone and is not having any value. But a determinant can be
reduced to a number,
(ii) The number of rows may not be equal to the number of columns in a
matrix. In a determinant the number of rows is always equal to the
number of columns.
15
(iii) On interchanging the rows and columns, a different matrix is formed.
In a determinant interchanging the rows and columns does not alter
the value of the determinant.
1.2.2 Definitions:
To every square matrix A of order n with entries as real or complex
numbers, we can associate a number called determinant of matrix A and it is
denoted by I A I or det (A) or A.
Thus determinant formed by the elements of A is said to be the determinant
of matrix A.
If A =
«11 «12
«21 «22.
then its I A I =
an «12
«21 «22
= an a 22 _ #21«12
To evaluate the determinant of order 3 or above we define minors and
cofactors.
Minors:
Let I A I =  [ajj]  be a determinant of order n. The minor of an arbitrary
element a,y is the determinant obtained by deleting the i row and_/ column in
which the element a„ stands. The minor of a„ is denoted by My.
Cofactors:
The cofactor is a signed minor. The cofactor of a,y is denoted by A y  and is
defined as Ay = ( 1) ; +j My.
The minors and cofactors of an, ai2, ^13 of a third order determinant
are as follows:
an
a\2
an
ail
an
an
031
«32
«33
an a 2 3
ail «33
(i) Minor of aj 1 is Mn =
Cofactor of an is An = (1)
(ii)Minor of «i2 is M[2 =
: a 22«33  a 32 «23
1 + 1
Mn =
an a 2 3
«32 «33
= «22«33 ~ a 32 <223
ail a 2 3
«31 «33
a 2i «33 «31«23
1+2,
Cofactor of a 12 is A 12 = (l) M 12 =
«21 «23
«31 «33
=  (a21«33  «23 «3l)
16
(iii) Minor of a 13 is Mi 3 =
a 2 \ an
«31 a 32
: «21 «32~«31 a 22
1 + 3
Cofactorof an is A13 = ( 1) M13 =
: «21 «32  «31 «22
«21 «22
«31 (232
Note: A determinant can be expanded using any row or column as given below:
a n
«12
013
a 2 \
<222
023
031
«32
«33
LetA =
A = an An + ai2 A12 + an A13 or a\\ Mn  ai2 M12 + ai3 M13
(expanding by Ri)
A = a n An + «2lA2l + ^31 A 31 or «11 M ll ~ «21 M 2 i + 031 M 31
(expanding by Ci)
A = a 2 \ A21 + (222 A22 + «23 A 23 or  (2 2 l M 2 i + «22 M 2 2  «23 M 23
(expanding by R2)
Example 1.8:
Find the minor and cofactor of each element of the determinant
3 4 1
012
52 6
Solution:
Minor of 3 is Mn
=
1
2
2
6
=6+4=2
Minor of 4 isMi2
=
2
5 6
= 010 =  10
Minor of 1 is M13
=
1
5 2
=0+5=5
Minor of is M21
=
4 1
2 6
= 24 + 2 = 26
Minor of  1 is M22
=
3 1
5 6
= 185 = 13
Minor of 2 is M23
=
3 A
5 
2
=  6  20 =  26
17
Minor of 5 is M31 =
Minor of  2 is M32 =
Minor of 6 is M33 =
Cof actor of 3 is Aj 1 =
Cof actor of 4 is A 12 =
Cofactor of 1 is A13 =
Cof actor of is A21 =
Cofactor of  1 is A22 =
Cofactor of 2 is A23 =
Cofactor of 5 is A3 1 =
Cofactor of  2 is A32 =
Cofactor of 6 is A33 =
Singular and nonsingular matrices:
4 1
1 2
3 1
2
3
=8+1=9
=60=6
4
1
l + l
1+2
1 + 3
2+1
2 + 2
2 + 3
3 + 1
3 + 2
3 + 3
=30=3
Mn = Mn = 2
Mi2 = M 12 =10
M 13 = Mi3 = 5
M 2 i =  M 2 i =  26
M22 = M22= 13
M23 =  M23 = 26
M 3 i=M 31 =9
M32 =  M32 =  6
M 33 = M33 =  3
A square matrix A is said to be singular if I A I =
A square matrix A is said to be nonsingular matrix, if I A I ^ 0.
is a singular matrix.
1 2
3"
For example, A =
4 5 6
7 8 9_
1 2 3
c
v IAI =
4 5 6
= 1
8
7 8 9
2
4 6
7 9
+ 3
4 5
7 8
= 1(45  48)  2 (36  42) + 3(32  35)
= 3 + 129 =
"1 7
5~
B =
2 6 3
L 4 8 9_
is a nonsingular matrix.
v 1
Bl =
1
2
4
7 5
6 3
8 9
= 1
6 3
8 9
7
2 3
4 9
+ 5
2 6
4 8
= 1(54  24)  7(18  12) + 5 (16  24)
= l(30)7(6) + 5(8)
=  52 *
.". The matrix B is a nonsingular matrix.
1.2.3 Properties of Determinants
There are many properties of determinants, which are very much useful in
solving problems. The following properties are true for determinants of any
order. But here we are going to prove the properties only for the determinant of
order 3.
Property 1:
The value of a determinant is unaltered by interchanging its rows and
columns.
Proof:
LetA =
Expanding A by the first row we get,
A = ai(b2C 3 b 3 C2)bi(a2C 3 a 3 C2) + c l (a2b 3 a 3 b2)
= a\b2Cj,  a\bj,C2  a.2b\c 3 + a 3^1 c 2 + a 2^3 c l ~~ a 3^2 c l
Cl\
b\
c\
«2
b 2
C2
«3
b 3
C3
Let us interchange the rows and columns of A.
determinant.
...(1)
Thus we get a new
A, =
a\ a.2
«3
b\ b 2
b 3
C[ c 2
C3
Since determinant can be expanded by any row or any
column we get
Ai = ai(&2C3  C2&3)  ^1 («2C3  C2a3) + ci(a2&3  M3)
= ai£>2 c 3 _ a \b^C2  02^1 c 3 + «3^lC2 + 02^3 C 1 _ 03^2 C 1 ■ • ■ (2)
From equations (1) and (2) wehaveA = A[ Hence the result.
Property 2:
If any two rows (columns) of a determinant are interchanged the
determinant changes its sign but its numerical value is unaltered.
Proof:
ci\ b\ c\
LetA =
«2
«3
b 2
b 3
19
A = a\(b2 Co,  bj C2)  bi(ci2 Co,  C13 C2) + c\ (a2bi  03 b2)
A = ai^2 c 3 _a 1^3 c 2 _a 2^1 c 3 + fl 3^1 c 2 + fl 2^3 c l _fl 3^2 c l •••(!)
Let Ai be the determinant obtained from A by interchanging the first and
second rows. i.e. R[ and R2.
ci2 b 2 c 2
Ai = a\ b\ c\
ai b 3 c 3
Now we have to show that Ai =  A.
Expanding Aj by R2, we have,
Ai = ai(b2C3b 3 C2)+b l (a2C3a 3 C2)ci(a2b 3 asb2)
= [aib2C3aibT,C2 + a 2 bic 3 + a 3 b l C2 + a2b i c l a3b2Ci] ...(2)
From (1) and (2) we get Ai =  A.
Similarly we can prove the result by interchanging any two columns.
Corollary:
The sign of a determinant changes or does not change according as there is
an odd or even number of interchanges among its rows (columns).
Property 3:
If two rows (columns) of a determinant are identical then the value of the
determinant is zero.
Proof:
Let A be the value of the determinant. Assume that the first two rows are
identical. By interchanging Rj and R2 we obtain  A (by property2). Since Rj
and R2 are identical even after the interchange we get the same A.
i.e. A = A => 2A = i.e. A =
Property 4:
If every element in a row (or column) of a determinant is multiplied by a
constant "fe" then the value of the determinant is multiplied by k.
Proof:
LetA =
Cl\
b\
c\
a 2
bi
C2
«3
b 3
C3
20
Let A j be the determinant obtained by multiplying the elements of the first
ka\ kb\ kci
row by 'fe' then Aj =
a 2 b2 c 2
a 3 b 3 c 3
Expanding along Ri we get,
Aj = ka\(b2C 3 b 3 C2)kb\(ci2C 3  a 3 C2) + kc\(fl2b 3  a 3 b2)
= fc[ai&2 c 3 ~~ a \b 3 C2 ~ a lb\c 3 + a 3 b\C2 + <22^3 C 1 ~~ a 3^2 c l]
Ai = kA. Hence the result.
Note:
(1) Let A be any square matrix of order n. Then kA is also a square matrix
which is obtained by multiplying every entry of the matrix A with the
scalar k. But the determinant k IAI means every entry in a row (or a
column) is multiplied by the scalar k.
(2) Let A be any square matrix of order n then I kA I = fe n l A I.
Deduction from properties (3) and (4)
If two rows (columns) of a determinant are proportional i.e. one row
(column) is a scalar multiple of other row (column) then its value is zero.
Property 5:
If every element in any row (column) can be expressed as the sum of two
quantities then given determinant can be expressed as the sum of two
determinants of the same order with the elements of the remaining rows
(columns) of both being the same.
ai+*i Pi+ji yi + zi
Proof: Let A = by b 2 b 3
C{ c 2 c 3
Expanding A along the first row, we get
A = (ai + x\)
b2 bi
ci c 3
(Pi+yi)
by 63
c\ c 3
+ (Yl + zi)
c\
in
= <M
b 2 b 3
C2 c 3
P
b 3
C3
+ Y1
+ Ui
in
C2
b 2 b 3
C2 c 3
b\ b 3
b\ b 2
y\
+ z\
c\ c 3
c\ c 2
21
xi y\
Z\
by b 2
h
c\ c 2
C3
ai pi yi
b\ b 2 b 3
c\ c 2 C3
Hence the result.
Note: If we wish to add (or merge) two determinants of the same order we add
corresponding entries of a particular row (column) provided the other entries in
rows (columns) are the same.
Property 6:
A determinant is unaltered when to each element of any row (column) is
added to those of several other rows (columns) multiplied respectively by
constant factors.
i.e. A determinant is unaltered when to each element of any row (column)
is added by the equimultiples of any parallel row (column).
Proof:
ci\ b\ c\
LetA= a 2 b 2 c 2
a 3 bi C3
Let Aj be a determinant obtained when to the elements of Ci of A are
added to those of second column and third column multiplied respectively by I
and m.
a\ + lb\ + mc\
A]= a 2 + lb 2 + mc 2
aj, + Ibj, + uct,
a\ b\ c\
a 2 b 2 c 2
ai h c 3
a\ b\ c\
a 2 b 2 c 2
aj, b 3 c 3
Therefore Aj = A.
01
b 2
b 3
\b {
c\
C2
C3
'l c\
lb 2 b 2 c 2
lbs h ct,
mc\
b\
c\
mc 2
b 2
C2
mci,
b 3
C3
(by property 5)
+ +
Ci is proportional to C 2 in the second det.
Ci is proportional to C3 in the third det.
Hence the result.
Note:
(1) Multiplying or dividing all entries of any one row (column) by the
same scalar is equivalent to multiplying or dividing the determinant
by the same scalar.
22
(2) If all the entries above or below the principal diagonal are zero (upper
triangular, lower triangular) then the value of the determinant is equal
to the product of the entries of the principal diagonal.
For example, let us consider
3 2 7
I A I = 5 3 =3(50) 2(00) + 7(00) = 15
1
The value of the determinant A is 15.
The product of the entries of the principal diagonal is 3 x 5 x 1 = 15.
x  1 x x 2
x2 x3 =0
x3
Solution: Since all the entries below the principal diagonal are zero, the value
of the determinant is (x  1) (x  2) (x  3)
.. (x l)(x2)(x3) = =^ x=l, x = 2, x = 3
x 5 1 1
=
36 =
Example 1.9: Solve
Example 1.10:
Solve for x if
x 5
7 x
+
1 2
1 1
Solution :
x 5
7 x
+
1 2
1 1
=
=> (x 2 35) + (l2) = => jc 2  35  1
=>
x 2 =
36
=> x =
i
6
Example 1.11: Solve for x if
Solution:
<°> 3 x  1 1 x + <°)
x +x = i.e. x(l — x) =
Example 1.12: Evaluate (i)
Solution:
(i) Let A =
1
A'
2
X
1
3
X
=
=
x = 0, x = 1
0l[x x] + =
a £> + c
b c + a
(ii)
c a + b
x + 2a x + 3a x + Aa
x + 3a x + Aa x + 5a
x + Aa x + 5a x + 6a
a
b + c
b
c + a
=
c
a + b
1 a a + b + c
1 b a + b + c
1 c a+b + c
c 3 ^ c 3 + c 2
23
= [v Ci is proportional to C3]
(ii) Let A =
=
Example 1.13:
Solution:
x + 2a
x + 3a
x + Aa
x + 2a
a
x + 3a
x + 4a
x + 5a
=
x+ 3a
a
x + Aa
x + 5a
x + 6a
x + Aa
a
2a
2a
2a
C 2 > C 2
c 3 > c 3
Ci
■Ci
[
.• C 2 is proportional
toC 3 ]
2x + y
x y
Prove that
2y + z
y z
=
2z + x
Z X
2x + y
x y
2x x y
y x y
2y + z
y z
—
2y y z
+
z y z
2z + x
Z X
2z z x
X z X
Ciis
proportional to C 2 in the first det."
J + u
'•'Ci
is
identical to 1
"3 in
the second det. _
=
Example 1.14: Prove that
Solution:
1
a
a 2
1
b
b 2
1
c
2
c
= (a b) (b  c) (c  a)
1
a
a 2
1
b
b 2
1
c
c 2
ab
a L b L
Ri
> Ri  R2
—
bc b 2 c 2
1 c c 2
R 2 > R2  R3
1 a + b
Take (a  b) and (b 
c)
= (ab)(b c)
1 b + c
from Ri and R 2
1 c
c 2
respectively.
Example 1.15:
Solution:
= (ab) (bc) [(1) (b + c) (1) (a + b)] = (ab) (bc) (ca)
■■xy
1
1
1
"ove that
1
l+x I
1
1 l+y
1 1
1 1 1
l+x I
=
1
I I
+ y
y
R 2 > R2
R3 > R3
Ri
Ri
= xy [ ".■ upper diagonal matrix]
24
Example 1.16: Prove that
Mb 2
Me 2
be
b + c
ca
c + a
ah
a + b
=
\lar
Mb 2
Mc 2
be b + c
ca c + a
ab a + b
1
abc
abc
abc
1
abc
1
abc
(ab+bc+ca)
abc
Ma abc a(b + c)
Mb abc b(c + a)
Mc abc c(a + b)
Ma 1 a(b + c)
Mb 1 b(c + a)
Mc 1 c(a + b)
be 1 a(b + c)
ca 1 b(c + a)
ab 1 c(a + b)
be 1 ab + be + ca
ca 1 ab + be + ca
ab 1 ab + be + ca
be 1 1
ca I I
ab 1 1
Multiply R 1; R 2 , R3
by a, b, c
respectively
Take abc from C2
Multiply Ci by abc
C 3 > C 3 + Ci
Example 1.17: Prove that
Solution: Let A =
(ab + be + ca) ,„ N
abc
=
W
,2 2 ,
C DC
b + c
5 that
2 2
c a ca
c + a
a b ab
a + b
^ 2
be b + c
c 2 « 2
ca c + a
« 2 ^ 2
ab a + b
Take (ab + bc + ca) from C3
[ v C2 is identical to C3]
=
Multiply Ri, R2 and R3 by a, o and c respectively
A =
a&c
2 2
ai> c abc ab + ac
2 2
be a abc be + ab
ca b abc ca + be
25
(abc)
abc
be 1 ab + ac
ca 1 be + ab
ab 1 ca + be
Take abc from Ci and C 2
be 1 ab + be + ca
= abc ca 1 ab + be + ca
ab 1 ab + be + ca
be 1 1
= abc (ab + be + ca) ca 1 1
ab 1 1
= abc (ab + be + ca) (0)
=
a+b+c c
C 3 > C 3 + Ci
Take (ab + be + ca) from C3
[ v C 2 is identical to C3]
Example 1.18 : Prove that
Solution:
a+b+c c b
c a+b+c a
—b —a a+b+c
c a+b+c a
b a a+b+c
= 2(a+b) (b+c) (c+a)
a + b a + b  (a + b)
 (b + c) b + c b + c
— b a a + b + c
Ri >Ri + R 2
R 2 > R2 + R 3
1
= (a + b) (b + c)
= (a + b)(b + c)
= (a + b)(b + c)x ( 2)
1 1
1 1 1
b a a+b+c
2
11 1
 b a a + b + c
1 1
b a+b+c
= (a + b)(b + c)x ( 2) [ (a + b + c) + b]
= (a + b)(b + c)x ( 2) [ a  c]
A = 2(a + b) (b + c) (c + a)
Take (a + b), (b + c)
from Ri and R 2
respectively
Ri > R t + R 2
Example 1.19: Prove that
2 , i
a + k
ab
ac
ab
b 2 + k
be
ac
be
c 2 + k
= k 2 (a 2 + b 2 + c 2 + k)
26
Solution: Let A =
a 2 + X
ab
ab
b 2 + X
cu
be
c + X
ac be
Multiply Ri, R2 and R3,by a, b and c respectively
A =
1
abc
a(a + X) a b
ab 2 b(b 2 + X)
2
a c
ac
be c(c + X)
Take a, b and c from Ci, C2 and C3 respectively
A =
abc
abc
a 2 + X
b 2
,2
b 2 + X
c 2 + X
a 2 +b 2 +c 2 +X a 2 +b 2 +c 2 +X a 2 +b 2 +c 2 +X
= (a 2 + b 2 + c 2 + X)
= (a 2 + b 2 + c 2 + X)
= (a 2 + b 2 + c 2 + X)
b 2 +X
c 2 +X
Ri> Ri + R 2 + R3
1 1 1
b 2 b 2 + X b 2
c 2 + X
c c
1
b 2 X
c 2 X
X
X
C 2 ^C 2 C!
C 3 ^C 3 C!
2 , 1
a + X
ab
ac
ab ac
b 2 + X be
be c 2 + X
= X\a 2 + b 2 + c 2 + X)
EXERCISE 1.2
2
(1) Find the value of the determinant
expansion.
15
10
without usual
27
(2) Identify the singular and nonsingular matrix
"1 4
9"
' 1
2
3 "
(i)
4 9 16
_9 16 25_
(ii)
4 5 6
_ 2  4 6.
2 jc 4
4 3 9
(3)
Solve (i)
3 2 1
1 2 3
= ~ 3 (ii)
32 7
4 4 jc
= 1
a  b b  c c  a
1 ab c(a + b)
(4)
Evaluate (i) b  c c  a a b
(ii)
1 be a(b + c)
c  a a  b b  c
1 ca b(c + a)
a  b  c 2a 2a
(5)
Prove that
2b b  c  a 2b
2c 2c c  a  b
= (a + b + c) 3
l + a 1 1
= abc(\+l + l + ^\
(6)
Prove that
1 \+b 1
1 1 1 + c
where a, b, c are non zero real numbers and hence evaluate the
l + a 1 1
value of
1 l+a 1
1 1 l + a
1 a a
(7)
Prove that
1 b b 3
1 c c 3
= (a  b) (b  c) (c  a) (a + b + c)
XX 1 — X
(8)
If x, y, z are all different and
2 1 3
y y \y
2 1 3
z z \z
=
then show that xyz = 1
1 2
1 a a
I a be
(9)
Prove that (i)
1 b b 2 
1 2
lcc
 \ b ca
1 c ab
y + Z x y
(ii)
Z + X Z X
= (x + y + z) (xzj 2
x + y
y z
28
(10) Prove that
b+c c+a a+b
q+r r+p p+q
y+z z+x x+y
' b c
• c a =3 a
a b
a b c
a  b b  c c  a
• + c c + a a + b
(i)
(iii)
(iv)
a b c
=2
p q r
x y z
(ii)
a ab ac
ab b be
2
ac be c
= 4a 2 b 2 c 2
= 3abc a —b — c
= a + b + c — 3abc
1.2.4 Factor method
Application of Remainder theorem to determinants
Theorem:
If each element of a determinant (A) is polynomial in x and if A vanishes
for x = a then (x  a) is a factor of A.
Proof:
Since the elements of A are polynomial in x, on expansion A will be a
polynomial function in x. (say p(x)). For x = a, A =
i.e. p(x) = when x = a, i.e. p(a) =
.". By Remainder theorem (x  a) is a factor of p(x).
i.e. (x  a) is a factor of A.
Note:
(1) This theorem is very much useful when we have to obtain the value of
the determinant in 'factors' form. Thus, for example if on putting
a = b in the determinant A any two of its rows or columns become
identical then A = and hence by the above theorem a  b will be a
factor of A.
(2) If r rows (column) are identical in a determinant of order n (n > r)
when we put x = a, then (x  a) r ~ is a factor of A.
(3) (x + a) is a factor of the polynomial fix) if and only if x =  a is a root
of the equation/(x) = 0.
29
Remark: In this section we deal certain problems with symmetric and cyclic
properties.
Example 1.20: Prove that
c c
= (a b) (b  c) (c  a) (a + b + c)
Solution:
1
a a
LetA =
1
b b 3
1
c c 3
.Puta = b, A =
1
1
lcc
= [ v Ri is identical to R2]
Thus
a a
b b 3
„3
= (a  b) (b  c) (c  a) k(a + b + c)
.'. (a  b) is a factor of A.
Similarly we observe that A is symmetric in a, b, c, by putting b = c, c = a,
we get A = 0. Hence (b  c) and (c  a) are also factors of A. .". The product
(ab) (bc) (c  a) is a factor of A. The degree of this product is 3. The product
3
of leading diagonal elements is 1 . b . c . The degree of this product is 4.
.". By cyclic and symmetric properties, the remaining symmetric factor of
first degree must be k(a + b + c), where k is any nonzero constant.
1
1
1 c c~
To find the value of k, give suitable values for a, b, c so that both sides do
not become zero. Take a = 0, b= 1, c = 2.
= fe(3)(l)(l)(2) => k=l
:. A = (a b) (b  c) (c a) (a + b + c)
Note: An important note regarding the remaining symmetric factor in the
factorisation of cyclic and symmetric expression in a, b and c
If m is the difference between the degree of the product of the factors
(found by the substitution) and the degree of the product of the leading diagonal
elements and if
(1) m is zero then the other symmetric factor is a constant (k)
(2) m is one then the other symmetric factor of degree 1 is k(a + b + c)
(3) m is two then the other symmetric factor of degree 2 is
2 2 2
kia + b + c )+l (ab+bc+ca)
1
1
1
1
1
2
8
30
Example 1.21:
Prove by factor method
Solution:
1 2 3
\ a a
1 2 3
= (a b) (b c) (ca) (ab + be + ca)
LetA =
2 3
I a a
1
a ui
1 c 2 c 3
Put a=b A =
1 c 2 c 3
= [vRi = R 2 ]
.". (a  b) is a factor of A.
By symmetry on putting b = c and c = a we can easily show that
A becomes zero and therefore (b  c) and (c  a) are also factors of A.
This means the product (a  b) (b  c) (c  a) is a factor of A. The degree
2 3
of this product is 3. The degree of the product of leading diagonal elements b c
is 5.
2 2 2
.". The other factor is k(a + b + c ) + l(ab + be + ca)
1 2 3
1 a a
1 b 2 b 3
1 c 2 c 3
= [k(a 2 + b 2 + c 2 ) + l(ab + be + ca)] (a  b) (b  c) (c  a)
To determine k and I give suitable values for a, b and c so that both sides
do not become zero. Take a = 0, b = 1 and c = 2
= [k (5) + 1(2)] ( 1) ( 1) (2)
1
1
1
1
1
4
8
4 = (5k + 21) 2
5k + 21 = 2
...(1)
Again put a = 0, b =  1 and c = 1
1
111 = [k(2) + /( 1)] (+1)(2)(1)
1 1 1
=> 2 = (2k  I) ( 2) ^ 2fc  Z =  1
On solving (1) and (2) we get k = and Z = 1
...(2)
31
.". A = (ab + be + ca) (a  b) (b  c) (c  a)
= (a  b) (b  c) (c  a) (ab + be + ca)
Example 1.22: Prove that
Solution:
(b + cf
u2
a
(c + a) 2
2
(a + by
= 2abc (a + b + c)~
Let A =
(b + cf
,2
A =
(b + c)
,2
(c + a)
2
Put a = we get
2 j2
c b
a
(a + by
= [ v C2 is porportional to C3]
c c
.'. (a  0) = a is a factor of A.
Similarly on putting b = 0, c = 0, we see that the value of A is zero.
.". a, b, c are factors of A. Put a + b + c = 0, we have
A =
(aY
b 2
2
(by
(cY
=
2 ■ .
Since three columns are identical, (a + b + c) is a factor of A.
2
.'. abc (a + b + c) is a factor of A and is of degree 5. The product of the
leading diagonal elements (b + c) (c + a) (a + b) is of degree 6.
.". The other factor of A must be k(a + b + c).
(b + c y
u2
(c + a)
2
= k abc (a + b + c)~
c c (a + b)
Take the values a = 1, b = 1 and c = 1
4 1 1
1 4 1
1 1 4
= k(l) (I) (I) (3)"
54 = 27k
k = 2
.. A = 2abc (a + b + c)~
32
Example 1.23: Show that
Solution:
Let A =
X
a
a
a
X
a
a
a
X
X
a
a
a
X
a
a
a
X
Put x = a
= (x  a) ix + 2d)
:. A =
\2
a
a
a
a
a
a
a
a
a
=
Since all the three rows are identical (x a) is a factor of A.
Put x =  2a.
■2a a a
A = a 2a a
a a —2a
(x + 2d) is a factor of A.
a a
2a a
a 2a
= [C 1 ^C 1 + C 2 + C 3 ]
.". (x a) (x + 2a) is a factor of A and is of degree 3. The degree of the
product of leading diagonal element is also 3. Therefore the other factor must be
k.
= k(x a) (x + 2a).
X
a
a
a
X
a
a
a
X
Equate x term on both sides, 1 = k
Example 1.24: Using factor method, prove
Solution:
Put;c=l, A =
Let A =
2 3 5
2 3 5
2 3 5
jc+1 3
2 x+2
2 3
=
A'
a
a
a
X
a
= (
a
a
X
x+l
3
5
2 x+2
5
2
3
x+4
5
5
A'
+ 4
= (x  a) ix + 2d)
= (xiy ( x + 9)
Since all the three rows are identical, (x  1) is a factor of A.
33
Put x = 9 in A, then A =
3 5
=
07 5
3 5
= [vC^d+Ci+Cg]
8 3 5
27 5
2 35
.'. (x + 9) is a factor of A.
The product (x  1) (x + 9) is a factor of A and is of degree 3. The degree
of the product of leading diagonal elements (x + 1) (x + 2) (x + 4) is also 3.
.". The remaining factor must be a constant "fe"
x+l 3 5
2 x + 2 5 = fc(x  l) 2 (x + 9). Equating x 3 term on both
2 3 jc + 4
sides we get k = 1
Thus A = (x  l) z (x + 9)
(1) Using factor method show that
(2) Prove by factor method
(3) Solve using factor method
EXERCISE 1.3
I a a
1
lcc
b + c a  c a  b
b  c c + a b  a
c  b c  a a + b
x + a b c
= (a  b) (b  c) (c  a)
= 8abc
a
a
x + b c
b x + c
=
(4) Factorise
■ 2 2
i c
(5) Show that
= (a + b + c) (a b) (b  c) (c  a)
be ca ab
b + c a a"
c + a
a + b c c z
1.2.5 Product of determinants
Rule for multiplication of two determinants is the same as the rule for
multiplication of two matrices.
While multiplying two matrices "rowbycolumn" rule alone can be
followed. The process of interchanging the rows and columns will not affect the
value of the determinant. Therefore we can also adopt the following procedures
for multiplication of two determinants.
34
(1) Rowbyrow multiplication rule
(2) Columnbycolumn multiplication rule
(3) Columnbyrow multiplication rule
Note: The determinant of the product matrix is equal to the product of the
individual determinant values of the square matrices of same order.
i.e. Let A and B be two square matrices of the same order.
We have I AB I = I A I I B I
This statement is verified in the following example.
Example 1.25: If A =
cos9  sin6
_sin6 cos8 .
,B =
cos8 sin8
 sin9 cos0.
are two square matrices
then show that I AB I = I A I I B I
Solution:
Given that A =
AB =
cos9  sin9
sin9 cos9
cos9  sin9
sin9 cos9
cos 2 9 + sin 2 G
andB =
cos9 sin9
 sin9 cos9.
cos9 sin9
 sin9 cos9.
cos9 sin0  sin0 cos9
IABI =
IAI =
IBI =
_sin0 cos0  cos0 sin0
1
cos + sin 6
1
cos9
sin9
cos9
 sin9
= 1
 sin9
cos9
sin9
cos9
= cos 2 9 + sin 2 9 = 1
= cos 2 9 + sin 2 9 = 1
IAI I B 1= 1x1 = 1
From (1) and (2)
Example 1.26: Show that
Solution: L.H.S. =
I AB I = I A I IBI
c
b
2
c
o
a
=
b
a
,2 , 2
7 + C
ab
ab
2 2
c + a
be
ac
be
a +1
c
b
2
c
b
c
b
c
a
=
c
a
c
a
b
a
o
b
a
b
a
o + c + b o + o + ab
2 2
o + o + ab c + o + a
o + ac + o be + o + o
o + ac + o
be + o + o
b 2 + a 2 +
(1)
(2)
35
2 , i2
c + b
ab
ac
ab
2 2
c +a
ac
be
*2 , 2
b + a
= R.H.S.
Example 1.27: Prove that
Solution:
L.H.S. =
a\ b\
a 2 b2
a\ b\
a 2 b2
a\ a 2
by b 2
a\ b\
02 b 2
a\ b\
a 2 b 2
be
2 2
a\ + <22 a\b\ + a 2 b 2
2 2
aj&l + (22^2 ^1 + i>2
a\ b\
a 2 b 2
/^Interchange rows and
Vcolumns of the first determinant
2 2
a\ + a 2 a\b\ + a 2 b 2
a\b\ + a 2 b 2 by + b 2
Example 1.28: Show that
Solution:
R.H.S. =
2bc — a
c 2
c
2ca  b
2
a
b 2
a 2
lab — c
a
b
c
=
b
c
a
c
a
b
a
b
c
2
a
b
c
a
b c
b
c
a
=
b
c
a
b
c a
c
a
b
c
a
b
c
a b
a
b
c
a
b c
=
b
c
a
x(
1)
c
a b
'
c
a
b
t
?
c a
Interchanging R2 and R3
in the 2 nd determinant
a b c
b c a
cab
 a +bc + cb
2
— ab + c + ab
, *2
 ac + ac + b
2bc  a
,2
 b  c
a b
c a
2
 ab + ab + c
2
 b + ac + ac
be + a + be
, j2
ac + b
+ ac
be + be
2
+ a
c + ab
+ ba
c z b z
2ac  b a
9 9
a lab  c
= L.H.S.
36
1.2.6 Relation between a determinant and its cofactor determinant
a\ b\ c\
Consider A = a 2 b 2 c 2
a?, bj, c 3
Let Ai, Bi, Ci be the cofactors of a\, b\,c\ in A
.". The cofactor determinant is
Ai
Bi C
l
t is
A 2 B 2 C 2
A3 B 3 C 3
.. A =
= Cl\
b 2 c 2
h c 3
■b\
a 2 c 2
«3 c 3
+ C\
a 2 b 2
aj, bj,
Let A be expanded by Ri
=> A = a\ (cofactor of a\) + b\ (cofactor of b\) + c\ (cofactor of c\)
=> A = a\A\ + by Bj + c\ <Z\
i.e. The sum of the products of the elements of any row of a
determinant with the corresponding row of cofactor determinant is equal
to the value of the determinant.
Similarly A = a 2 A 2 + b 2 B 2 + c 2 C 2 A = C13A3 + & 3 B 3 + c 3 C 3
Now let us consider the sum of the product of first row elements with the
corresponding second row elements of cofactor determinant i.e. let us consider
the expression
aiA 2 + &iB 2 + ciC 2
=  a\
b\
c\
a\
c\
a\
b\
b 3
ci
+ h
a 3
C3
c\
«3
b 3
asb\)
=  ai(biCT,  b^cx) + b\(a\c?,  a 3 ci)  c\(a\b?, ■
=
.". The expression a\A 2 + b\B 2 + c\C 2 =
Thus we have
aiA 3 + ^iB 3 + C1C3 = ; a 2 Ai + b 2 B[ + c 2 C\ = ; a 2 A 3 + & 2 B 3 + c 2 C 3 =
a 3 Ai +£ 3 Bi +c 3 Ci =0 ; a 3 A 2 + £ 3 B 2 + c 3 C 2 =
i.e. The sum of the products of the elements of any row of a
determinant with any other row of cofactor determinant is equal to
Note: Instead of rows, if we take columns we get the same results.
.". A = a[A\ + a 2 A 2 + a 3 A 3
A = friBi + £> 2 B 2 + £ 3 B 3
A = ciCi + c 2 C 2 + c 3 C 3
Thus the above results can be put in a tabular column as shown below.
37
Rowwise
Columnwise
Ri
R 2
R 3
Ci
c 2
c 3
n
A
c\
A
n
A
ci
A
n
A
C3
A
a\ b\
c\
a 2 b 2
ci
«3 h
Ci
Where r,'s q's are i row and i column of the original determinant R/s, Q's
are i row and i column respectively of the corresponding cofactor determinant.
Example 1.29: If Ai , B i, Ci are the cofactors of a\, b\, c\ in A =
Ai Bi Ci
then show that A 2 B 2 C 2 = A z
A 3 B 3 C 3
Ai Bi Ci
Solution: «2 ^2 c 2 A 2 B 2 C 2
A 3 B 3 C 3
a\A\ + b\B[ + c\C\ «iA 2 + i>iB 2 + cjC 2 aiA 3 + £>iB 3 + cjC 3
a 2 Ai + i> 2 Bj + c 2 Ci a 2 A 2 + £> 2 B 2 + cjC 2 a 2 A 3 + £> 2 B 3 + c 2 C 3
a 3 Ai + & 3 Bi + c 3 Ci a 3 A 2 + 6 3 B 2 + c 3 C 2 a 3 A 3 + 6 3 B 3 + c 3 C 3
A
a\ b\ c\
a 2 b 2 c 2
a 3 bj, c 3
A
A
= A J
i.e. A x
A]
Bi Ci
A 2
B 2 C 2
A 3
B 3 C 3
= A J
A,
Bi
Ci
A 2
B 2
c 2
A 3
B 3
c 3
= A Z
EXERCISE 1.4
1 a
a
2
, ~ 2 2 2
1  2a  a  a
(1) Show that
a 1
a
—
 a  1 a 2a
a a
1
2 2 ,
 a a —2a — 1
1 X
2
A'
a 1 2a
(a — jc) (£> — x)
(2) Show that
1 y
y 2
b 2 1 2b
=
(ay) 2 (by) 2
1 z
2
Z
c
2 1 2c
2
(az) (b*
:f
(cx)
(cy) 2
(cz) 2
38
2. VECTOR ALGEBRA
2.1 Introduction:
The development of the concept of vectors was influenced by the works of
the German Mathematician H.G. Grassmann (1809  1877) and the Irish
mathematician W.R. Hamilton (1805  1865). It is interesting to note that both
were linguists, being specialists in Sanskrit literature. While Hamilton occupied
high positions, Grassman was a secondary school teacher.
The best features of Quaternion Calculus and Cartesian Geometry were
united, largely through the efforts of the American Mathematician J.B. Gibbs
(1839  1903) and Q. Heariside (1850  1925) of England and new subject
called Vector Algebra was created. The term vectors was due to Hamilton and it
was derived from the Latin word 'to carry'. The theory of vectors was also
based on Grassman' s theory of extension.
It was soon realised that vectors would be the ideal tools for the fruitful
study of many ideas in geometry and physics. Vector algebra is widely used in
the study of certain type of problems in Geometry, Mechanics, Engineering and
other branches of Applied Mathematics.
Physical quantities are divided into two categories  scalar quantities and
vector quantities.
Definitions:
Scalar : A quantity having only magnitude is called a scalar. It is not
related to any fixed direction in space.
Examples : mass, volume, density, work, temperature,
distance, area, real numbers etc.
To represent a scalar quantity, we assign a real number to it, which gives
its magnitude in terms of a certain basic unit of a quantity. Throughout this
chapter, by scalars we shall mean real numbers. Normally, scalars are denoted
by a, b,c...
Vector : A quantity having both magnitude and direction is called a
vector.
Examples : displacement, velocity, acceleration, momentum,
force, moment of a force, weight etc.
Representation of vectors:
Vectors are represented by directed line segments such that the length of
the line segment is the magnitude of the vector and the direction of arrow
marked at one end denotes the direction of the vector.
39
A vector denoted by a = AB is
determined by two points A, B such that the
magnitude of the vector is the length of the
A
Fig. 2. 1
line segment AB and its direction is that from A to B. The point A is called
initial point of the vector AB and B is called the terminal point. Vectors are
> > »
generally denoted by a , b , c ... (read as vector a, vector b, vector c, ...)
Magnitude of a vector
> — >
The modulus or magnitude of a vector a = AB is a positive number
which is a measure of its length and is denoted by I a I =1 AB I = AB The
modulus of a is also written as 'a'
Thus
ul = a ; Im =fc ; l~cl =
I AB I = AB ; I CD I = CD ; I PQ I = PQ
Caution: The two end points A and B are not interchangeable.
Note: Every vector AB has three characteristics:
Length : The length of AB will be denoted by I AB I or AB.
Support : The line of unlimited length of which AB is a segment is
called the support of the vector AB ,
Sense : The sense of AB is from A to B and that of BA is from B to
A. Thus the sense of a directed line segment is from its initial
point to the terminal point.
Equality of vectors:
» » » — »
Two vectors a and b are said to be equal, written as a = b , if they
have the
(i) same magnitude (ii) same direction
40
In fig (2.2) AB II CD and AB = CD
AB and CD are in the same direction
> — > > >
Hence AB = CD or a = b
2.2 Types of Vectors
Zero or Null Vector:
A vector whose initial and terminal points are coincident is called a zero or
>
null or a void vector. The zero vector is denoted by O
Vectors other than the null vector are called proper vectors.
Unit vector:
A vector whose modulus is unity, is called a unit vector.
The unit vector in the direction of a is denoted by a (read as 'a cap').
Thus \a\ = 1
The unit vectors parallel to a are + a
ult: a
direction)]
> » A
Result: a = \ a \ a [i.e. any vector = (its modulus) x (unit vector in that
Itl
; ("a *o)
In general
vector in that direction
unit vector in any direction = modulus of the vector
Like and unlike vectors:
Vectors are said to be like when they have the same sense of direction and
unlike when they have opposite directions.
a c
like vectors
unlike vectors
Fig. 2. 3
41
Coinitial vectors:
Vectors having the same initial point are called coinitial vectors.
Coterminus vectors:
Vectors having the same terminal point are called coterminus vectors.
Collinear or Parallel vectors:
Vectors are said to be collinear or parallel if they have the same line of
action or have the lines of action parallel to one another.
Coplanar vectors:
Vectors are said to be coplanar if they are parallel to the same plane or they
lie in the same plane.
Negative vector:
»
The vector which has the same magnitude as that of a but opposite
> > — > >
direction is called the negative of a and is denoted by  a . Thus if AB = a
> >
then BA =  a .
Reciprocal of a vector:
Let a be a nonzero vector. The vector which has the same direction as
that of a but has magnitude reciprocal to that of a is called the reciprocal of
a and is written as V, a ) where \\a )
111
~~ a
Free and localised vector:
When we are at liberty to choose the origin of the vector at any point, then
it is said to be a free vector. But when it is restricted to a certain specified point,
then the vector is said to be localised vector.
2.3 Operations on vectors:
2.3.1 Addition of vectors:
Let OA = a , AB = b Join OB.
Then OB represents the addition (sum) of the
vectors a and b .
This is written as OA + AB = OB
Thus OB = OA + AB = a + b
42
This is known as the triangle law of addition of vectors which states that, if
two vectors are represented in magnitude and direction by the two sides of a
triangle taken in the same order, then their sum is represented by the third side
taken in the reverse order.
Applying the triangle law of addition of vectors in A
AABC, we have
BC + CA = BA
=> BC + CA =  AB
B C
=> AB+BC+CA="o' Fig. 2. 5
Thus the sum of the vectors representing the sides of a triangle taken in
order is the null vector.
Parallelogram law of addition of vectors:
a
>
If two vectors a and b are represented in
magnitude and direction by the two adjacent sides
c
of a parallelogram, then their sum c is
represented by the diagonal of the parallelogram
which is coinitial with the given vectors.
Symbolically we have "OP + OQ =OR r .. , ,
Fig. 2. 6
Thus if the vectors are represented by two adjacent sides of a
parallelogram, the diagonal of the parallelogram will represent the sum of the
vectors.
By repeated use of the triangle law we can find the sum of any number of
vectors.
> >
Let OA = a .
»
BC =~c .
CD = d
AB =
be any five vectors as shown in the fig (2.7). We
observe from the figure that each new vector is
drawn from the terminal point of its previous one.
OA + AB + BC + CD + DE = OE
Thus the line joining the initial point of the
first vector to the terminal point of the last vector is
the sum of all the vectors. This is called the polygon
law of addition of vectors.
43
Note : It should be noted that the magnitude of a + b is not equal to the sum
> >
of the magnitudes of a and b .
2.3.2 Subtraction of vectors:
If a and & are two vectors, then the subtraction of & from a is
> > > >
defined as the vector sum of a and  b and is denoted by a  o .
a — b = a
♦ (*)
Let
OA = a and AB =
>•
Then OB = OA + AB = a + b
To subtract & from a , produce BA to B'
such that AB = AB'. .. AF? =  AB =  b
a — b
Now by the triangle law of addition
OB' = OA + AB' = a + \ b ) =
Properties of addition of vectors:
Theorem 2.1:
> >
Vector addition is commutative i.e., if a and b are any two vectors then
» » » »
> > > >
Let OA = a AB = b
C t B
InAOAB, OA + AB = OB
rJ
' /I
(by triangle law of add.)
7
s^ Jr
=^> a + b = OB ... (1)
P
, /
Complete the parallelogram OABC
O
7J* A
CB = OA ="? ; OC = AB =~?
(J
Fig. 2. 9
In AOCB, we have OC + CB = OB
i.e. => b
+ "a =OB ...(2)
> » » »
From (1) and (2) we have a + £ = & + a
.". Vector addition is commutative.
44
Theorem 2.2:
Vector addition is associative
> » >
i.e. For any three vectors a , b , c
Proof :
> >
Let OA = a
Join O and B
In AOAB,
In AOBC,
In AABC,
In AOAC
AB =
OandC
>
BC = c
AandC
OA + AB =
» >
a + b =
OB + BC =
/> A >
AB + BC =
ft + c =
OB
OB
OC
oc
AC
AC
OC
OC
(1)
(2) [using (1)]
(3)
(4) [using (3)]
OA + AC =
> (> >^
=> a+^£+c,/ =
From (2) and (4), we have \a + b ) + c = a + \b + c )
:. vector addition is associative.
Theorem 2.3:
> » » — » — » » >
For every vector a, a +0=0+a=a where O is the null
vector, [existence of additive identity]
Proof:
Let OA = a
Then
Also
a + O = OA + AA = OA = a
.". a + O = a
O + a =00 + OA = OA = a
45
.". O + a = a
> » » » >
.". a+0=0+a=a
Theorem 2.4:
> >
For every vector a , there corresponds a vector  a such that
a +^ a ) = O =\— a )+a [existence of additive inverse]
Proof: LetOA = a . ThenAO = a
> / A > > > >
.. a + ^ a ^ = OA + AO = OO = O
(""«) +
a = AO + OA = AA = O
Hence a + ^ a / =^a/+a =0
2.3.3 Multiplication of a vector by a scalar
Let m be a scalar and a be any vector, then ma is defined as a vector
having the same support as that of a such that its magnitude is I m I times the
magnitude of a and its direction is same as or opposite to the direction of a
according as m is positive or negative.
Result : Two vectors a and b are collinear or parallel if and only if a = mb
for some nonzero scalar m.
»
For any vector a we define the following:
>
»
>
>
>
— >
(1) a
= a ;
( 1) a =
 a
; Oa
= O
Note: If a is a vector then 5 a is a vector whose magnitude is 5 times the
> > »
magnitude of a and whose direction is same as that of a . But 5a is a
vector whose magnitude is 5 times the magnitude of a and whose direction is
>
opposite to a .
Properties of Multiplication of vectors by a scalar
The following are properties of multiplication of vectors by scalars.
> >
For vectors a , b and scalars m, n we have
46
(i) wz\— a ) =(—m) a =  \m a )
(iii) ot^m a ) = (mn) a = n\m a )
(ii) (m)
\— a ) = m a
> > »
(iv) (m + n) a =ma + n a
Theorem 2.5 (Without Proof)
If a and b are any two vectors and m is a scalar
then wz\ a + b ) =ma + mb .
>
/> A
Result : m^ a  b ) = i
2.4 Position vector
If a point O is fixed as the origin in space
(or plane) and P is any point, then OP is
called the position vector (P.V.) of P with
respect to O.
> >
From the diagram OP = r
Similarly OA is called the position
vector (P.V.) of A with respect to O and OB
is the P.V. of B with respect to O.
Fig. 2.11
Theorem 2.6: AB = OB
respectively.
OA where OA and OB are the P. Vs of A and B
> »
Proof: Let O be the origin. Let a and b be the position vectors of points
A and B respectively
Then OA = a ; OB = b
In AOAB, we have by triangle law of
addition
OA + AB = OB
=> AB = OB
i.e. AB = (P.V of B) (P.V of A)
— > > >
OA = b  a
Fig. 2. 12
47
Note : In AB , the point B is the head of the vector and A is the tail of the
vector.
.. AB = (P. V. of the head)  (P. V. of the tail). Similarly BA = OA  OB
The above rule will be very much useful in doing problems.
Theorem 2.7: [Section Formula  Internal Division]
> >
Let A and B be two points with position vectors a and b respectively
and let P be a point dividing AB internally in the ratio m : n. Then the position
vector of P is given by
OP =
> »
n a + mb
m + n
Proof:
Let O be the origin.
Then OA = a
OB =
>
> > >
Let the position vector of P with respect to O be r i.e. OP = r
Let P divide AB internally in the ratio m : n
AP m
Then
PB _ n
n AP = m PB
m(oPOa) =m(oBOp)
n r  n a = mb — m r
» » >
(m + n) r = mb + n a
n AP = m PB
=> n \ r  a J =m\b — .
> > » >
=> m r + n r =mb + n a
r =
mb + n a
m + n
Result (1): If P is the mid point of AB, then it divides AB in the ratio 1:1.
.. The P.V. of P is
\ . b + \. a a + b
1 + 1
48
.. P.V. of the mid point P of AB is OP = r = ^ —
Result (2): Condition that three points may be collinear
Proof: Assume that the points A, P and B (whose P.Vs are a , r and b
respectively) are collinear
We have
> >
— » mb + n a
r = ;
m + n
> » — »
(m + n) r = mb + n a
(m + n) r
■Bfl =0
In this vector equation, sum of the scalar coefficients in the
L.H.S. =(m + n)mn =
Thus we have the result, if A, B, C are collinear points with position
> > »
vectors a , b , c respectively then there exists scalars x, y, z such that
> > > >
jfl +yb + zc =0 and x + y + z =
Conversely if the scalars x, y, z are such that x + y + z = and
» > » » > » >
jfl +y^ +zc =0 then the points with position vectors a , £ and c
are collinear.
Result 3: [Section formula  External division]
> >
Let A and B be two points with position vectors a and b respectively
and let P be a point dividing AB externally in the ratio m : n. Then the position
vector of P is given by
»
>
> m b
OP =
m
n a
 n
Proof:
Let be the
origin. A and B are the two
points whose position '
i
/ectors are a
>
and ^
Then OA =
= a
; OB =
Let P divide AB externally in the ratio m : n. Let the position vector of P
> > >
with respect to O be r i.e. OP = r
49
w u AP m
We have pb = 7
n AP =  m PB
^> «AP = mPB
ap & ft?
are in the opposite direction.
(cJPOa) = m(oBCJ>p) =^> n\r ~a) = m\r ~b)
» > > >
m &  « a = m r —nr
> — » — » >
=> n r — n a = m r — mb
» > >
=> mb — n a = (m — ri) r
» >
— > mb —n a
r =
m  «
Theorem 2.8: The medians of a triangle are concurrent.
Proof:
Let ABC be a triangle and let D, E, F be the mid points of its sides BC, CA
and AB respectively. We have to prove that the medians AD, BE, CF are
concurrent.
» > »
Let O be the origin and a , b , c be the position vectors of A, B, C
respectively.
The position vectors of D, E, F are
b + c c + a a + b
Affl)
2 ' 2 ' 2
Let Gj be the point on AD dividing it
internally in the ratio 2 : 1
B($) D C(Q
F/g. 2. 75
P.V. ofGj =
2QD + 1QA
2+ 1
+ c I ,>
+ 1 fl
OG, =
V
a + o + c
'l  3 =3
Let G 2 be the point on BE dividing it internally in the ratio 2 : 1
2QE +1QB
(1)
OGo =
2+ 1
50
c + a
+ 1. b > :> >
a + b + c
OG 2 = 3 = 3 ( 2 )
Similarly if G3 divides CF in the ratio 2 : 1 then
> a + b + c
OG 3 = 3 (3 )
From (1), (2), (3) we find that the position vectors of the three points
Gj, G 2 , G 3 are one and the same. Hence they are not different points. Let the
common point be denoted by G.
Therefore the three medians are concurrent and the point of
concurrence is G.
Result:
The point of intersection of the three medians of a triangle is called the
centroid of the triangle.
> a + b + c
The position vector of the centroid G of A ABC is OG = 3
» » >
where a , b , c are the position vectors of the vertices A, B, C respectively
and O is the origin of reference.
» — » — »
Example 2.1: If a , b , c be the vectors represented by the three sides of a
» — » > — »
triangle, taken in order, then prove that a + b + c = O
Solution:
Let ABC be a triangle such that
FiC = a , CA = b and AB = c
a + b + c = FiC + CA + AB
= BA + AB (.. BC + CA = BA
Example 2.2:
If a anc
hexagon, find the vectors determined by the other sides taken in order.
= BB =0
If a and b are the vectors determined by two adjacent sides of a regular
51
Solution:
Let ABCDEF be a regular hexagon
> > > >
such that AB = a and BC = b
Since AD II BC such that AD = 2.BC
AD = 2BC =Tb
In AABC, we have AB + ISC = AC
> > >
=> AC = a + b
InAACD, AD = AC + CD
Fig. 2. 17
> > — > > /> A
CD = AD  AC = 2 fc  V a + fe J =
DE =  AB =
FiF =  BC =
FA= CD =\b  a) =
Example 2.3:
b — a
a
^>
a — b
» — » > >
The position vectors of the points A, B, C, D are a , i , 2 a + 3o .
> > > — >
a 26 respectively. Find DB and AC
Solution: Given that
OA = a ; OB = b
; OC =2a + 36 ; OD
= a 
>
26
DB = OB OD = ~b 
^a 2fe^= Z? a +2b
= 36
 a
AC = OC OA
/> A >
= \2a + 36/  a
= a + 36
Example 2.4: Find the position vector of the points which divide the join of the
»
points A and B whose P.Vs are a
externally in the ratio 3 : 2
2 6 and 2 a
b internally and
52
Solution:
OA = a 26 ; OB =2a  b
Let P divide AB internally in the ratio 3 : 2
3 OB + 2QA 3(2"^  t) + 2\a  it)
P.V. ofP =
3 + 2
> » » > » »
6a 36 +2a 46 8a 76 8 > 7 >
' = c « — T b
5 5
Let Q divide AB externally in the ratio 3 : 2
( > A /'> V
^2 g  6 j_
n,, ^ 3QB2QA 3^ 6^2^ 2 ,
P.V. of Q = — = l
» > > » » »
= 6a 36 2a +46 =4a +6
Example 2.5: If a and 6 are position vectors of points A and B respectively,
then find the position vector of points of trisection of AB.
Solution:
Let P and Q be the points of
trisection of AB .*"" J £ ^ /E >,
Let AP = PQ = QB = X (say) ^ 2 7S
P divides AB in the ratio 1:2
P.V.ofP = O? = LM±OA = Ll±2^ = l±2l
Q is the midpoint of PB
» » » » »
6 +2a » 6 +2a +36
OP + OB 3 + * 3 2a +46
P.V. of Q = 2"
> >
g +26
3
Example 2.6: By using vectors, prove that a quadrilateral is a parallelogram if
and only if the diagonals bisect each other.
53
Solution:
Let ABCD be a quadrilateral q^/j c(r)
First we assume that ABCD is a parallelogram
To prove that its diagonals bisect each other
Let O be the origin of reference.
> > > > > > >
.. OA = a , OB = & , OC = c , OD = d &W B$
> > Fig. 2. 79
Since ABCD is a parallelogram AB = DC
=^> OB OA = OC OD =^> b  a = c  d
> » — » — »
> » > > b + d a + c
=> o + a = a + c => ~ = ~
i.e. P.V. of the midpoint of BD = P.V. of the midpoint of AC. Thus, the
point, which bisects AC also, bisects BD. Hence the diagonals of a
parallelogram ABCD bisect each other.
Conversely suppose that ABCD is a quadrilateral such that its diagonals
bisect each other. To prove that it is a parallelogram.
> > > »
Let a , b , c , d be the position vectors of its vertices A, B, C and D
respectively. Since diagonals AC and BD bisect each other.
P.V. of the midpoint of AC = P.V. of the midpoint of BD
•••(I)
=>
» — » > »
a + c b + d
2 _ 2
> > » »
^>a+c = b+d
=>
> — » » »
£>  a = c a
i.e. AB = DC
Also(l) =>
» > > >
d — a = c — b
i.e. AD = FiC
Hence ABCD
is a parallelogram.
•nple 2.7:
In a triangle ABC if D and E are the midpoints of sides AB and AC
respectively, show that BE + DC = y BC
54
Solution:
For convenience we choose A as the origin.
»
Let the position vectors of B and C be b and
»
c respectively. Since D and E are the
midpoints of AB and AC, the position vectors
>
c
of D and E are it and it respectively.
B {$) C(c)
Fig. 2. 20
Now
BE = P.V. ofEP.V. ofB =
c >
DC = P.V. of C P.V. of D = ~c y
— > > c > > b
BE +DC =^r6 +C
2 u r c 2
iC?*)
= j [P.V. of C P.V. of B]
= 2 BC
Example 2.8: Prove that the line segment joining the midpoints of two sides of
a triangle is parallel to the third side and equal to half of it.
Solution:
Let ABC be a triangle, and let O be the A (a)
origin of reference. Let D and E be the
midpoints of AB and AC respectively.
Let
OA = a , OB = b , OC = c
P.V. ofD=OD =
P.V. of E = OE =
a
+
2
a
+
c
B(b)
Now
DE = OE  OD =
a + c
a +
C05
Fig. 2. 27
55
Also
a + c 
2
afc 1 f»
= 2 V c "
DE
= i bc
=> DE II BC
DE
4^
=> Ide
■) 4 (
oc
Ob) =\ BC
DE I = o I BC I => DE = o BC
1
Hence DE II BC and DE = ^ BC.
Example 2.9: Using vector method, prove that the line segments joining the
midpoints of the adjacent sides of a quadrilateral taken in order form a
parallelogram.
Solution:
Let ABCD be a quadrilateral and let P, Q,
R, S be the midpoints of the sides AB, BC,
CD and DA respectively.
Then the position vectors of P, Q, R, S are
a + b b + c c + d d + a
D(d)
respectively.
z z z Ma) p B(b)
Fig. 2. 22
In order to prove that PQRS is a parallelogram it is sufficient to show that
PQ =^R and r^ = QR
Now PQ = P.V. ofQP.V. ofP =
SR = P.V. of R P.V. of S =
b + c
a + b
I 2 )~
V 2 J
c + d
d + a
{ 2 j
v 2 J
> >
c — a
c  a
.. PQ = SR
=^PQ II SR and PQ = SR
Similarly we can prove that PS = QR and PS II QR
Hence PQRS is a parallelogram.
Example 2.10 :
—> > >
Let a , b , c be the position vectors of three distinct points A, B, C. If
> — » — »
there exists scalars Z, m, n (not all zero) such that Z a +m /? +n c = and Z + m
+ n = then show that A, B and C lie on a line.
56
Solution:
It is given that I, m, n are not all zero. So, let n be a nonzero scalar.
»»—» > / > A
Z a + mb + n c = => »c =—\la+mb)
( > *\ /> >\ _> _>
> ya +mb ) > y a +mb ) la +mb
c = ~ n => c = "  (/ + m) = / + m
=> The point C divides the line joining A and B in the ratio m : Z
Hence A, B and C lies on the same line.
» — » > > > >
Note : a , b are collinear vectors=> a =Xb or b =Xa for some scalar A,
Collinear points: If A, B, C are three points in a plane such that AB = XBC
or AB = ^AC (or) BC = XAC for some scalar X, then A, B, C are
collinear.
Example 2.11: Show that the points with position vectors
» — » — » » » > > — » »
a 26 +3c ,2a +3/?  c and A a lb +1 c are collinear.
Solution:
Let A, B, C be the points with position vectors
> > — » — » — » — » » — » — »
a 2/? + 3 c ,  2 a +3/?  c and 4a 7/? + 7 c respectively.
— > >>> — > » > > — > » >• >
OA = a 2 b + 3c , OB =2 a +3b  c , OC = 4 a 7ft + 7 c
> > > / > > A /> > A
AB =OBOA =^2a+3/?cyVa2ft+3cJ
» — » — » — » — » — » » — » — »
= — 2a + 3 ft  c a +2 b 3 c =  3 a + 5 b Ac
> > > / > > A / > > A
BC=OCOB =V4a7ft+7cJV2a+3ftc y ;
> > — » — » — » — » > » — »
= 4a 7ft +1 c +2 a 3ft + c = 6a 10ft + 8 c
Clearly BC =6~a 10b + 8"? =  2 { 3~a +5~b  A~c) = 2(ab)
=> AB and BC are parallel vectors but B is a point common to them.
So AB and BC are collinear vectors. Hence A, B, C are collinear points.
57
EXERCISE 2.1
> > — > — >
(1) If a and b represent two adjacent sides AB and BC respectively of
a paralleogram ABCD. Find the diagonals AC and BD .
(2) If PO + OQ = QO + OR , show that the points P, Q, R are collinear.
(3) Show that the points with position vectors
> > — » » — » » » — »
a 2fc +3c ,2a + 3 b + 2 c and 8 a + 13 & are collinear.
> > — »
(4) Show that the points A, B, C with position vectors 2 a + 3 b + 5 c ,
> — » — » > >
a + 2 b + 3 c and 1 a  c respectively, are collinear.
(5) If D is the midpoint of the side BC of a triangle ABC, prove that
AB + AC =2 AD
(6) If G is the centroid of a triangle ABC, prove that GA +GB +GC = O
(7) If ABC and A'B'C are two triangles and G, G' be their corresponding
centroids, prove that AA* + BET + CC = 3GC?
(8) Prove that the sum of the vectors directed from the vertices to the
midpoints of opposite sides of a triangle is zero
(9) Prove by vector method that the line segment joining the midpoints of
the diagonals of a trapezium is parallel to the parallel sides and equal to
half of their difference.
(10) Prove by vector method that the internal bisectors of the angles of a
triangle are concurrent.
(11) Prove using vectors the midpoints of two opposite sides of a
quadrilateral and the midpoints of the diagonals are the vertices of a
parallelogram.
(12) If ABCD is a quadrilateral and E and F are the midpoints of AC and
BD respectively, prove that AB + AD + CB + CD = 4 EF
2.5 Resolution of a Vector
Theorem 2.9 (Without Proof) :
» > >
Let a and b be two noncollinear vectors and r be a vector coplanar
> >>>
with them. Then r can be expressed uniquely as r = I a + mb where I, m
are scalars.
58
Note : We call Z a + m b as a linear combination of vectors a and /? , where
/, m are scalars.
Rectangular resolution of a vector in two dimension
Theorem 2.10 :
If P is a point in a two dimensional plane which has coordinates (x, y)
> > > > >
then OP =ii + y _/' , where i and j are unit vectors along OX and
OY respectively.
Proof:
Let P(x, y) be a point in a plane with
reference to OX and OY as
coordinate axes as shown in the
figure.
Draw PL perpendicular to OX.
Then OL = x and LP = y
> >
Let i , j be the unit vectors along
OX and OY respectively.
Then OL = x i and X? =yj
Y
*
P(x, y)
J* i\
—>
i
X
Fig. 2. 23
Vectors OL and LP are known as the components of OP along xaxis
and yaxis respectively.
Now by triangle law of addition
"OP = "OL + X? = x i +yj = r (say)
r
= x i + y y
Now OP 2 = OL 2 + LP 2 = x 2 + y 2
=5> OP = V?+/ z* l7l = V* 2 + /
Thus, if a point P in a plane has coordinates (x, y) then
> > > >
(i) r = OP = x i + y j
(ii) I~r1 =IopI =lx7'ly7 > l =Vx 2 7y 2
> >
(iii) The component of OP along xaxis is a vector x i and the
— > >
component of OP along yaxis is a vector y 7
59
Components of a vector AB in terms of coordinates of A and B
Let A(x 1? yj) and B(x 2 , y 2 ) be any two
points in XOY plane. Let i and j be
unit vectors along OX and OY
respectively.
AN = x 2 Xj, BN=y 2 y 1
.. AN = (x 2  jci) i , NB
>
Now by triangle law of addition
Y"
A
Bfeyi)
/
7* >v
A(*i,yi)
O
]
L
M X
Fjg. 2. 24
>
»
AB = AN + NB ={x 2 x l )i + (y 2 y{) j
>
Component of AB along xaxis = (x 2  Xj) z
»
Component of AB along yaxis = (y 2 .yi) j
AB 2 = AN 2 + NB 2 = (x 2  x,) 2 + (y 2  y{) 2
=> AB =^(x 2 x 1 ) 2 + (y 2 v 1 ) 2
which gives the distance between A and B.
Addition, Subtraction, Multiplication of a vector by a scalar and equality
of vectors in terms of components:
Let
a
»
= aj i + a 2 j and £> = byi + b 2 j
We define
(i) a + £> = i^aj i + a 2 jj + \b[ i +b 2 j) = (aj + b{) i +(a 2 + b 2 )j
(ii) a — b = [a^ i +a 2 j
V +^2^7 =(«i»i) * +(a2 b 2)J
(iii) ma = m ^aj i +a 2 j J =ma\ i +ma 2 j
where m is a scalar
a = b
» > > >
aj z +a 2 j = i»j j +b 2 j
(iv) a = £> => aj i +a 2 j = b\ i +b 2 j => aj = £>janda 2 = £> 2
Example 2.12: Let O be the origin and P(— 2, 4) be a point in the xyplane.
> > > I >
Express OP in terms of vectors i and j . Also find I OP I
60
Solution: The position vector of P, OP = 2 i +4j
l~OP I = l2"f + 4"7l = \j( 2) 2 + (4) 2 =^4+16 =a/20
= 2^5
Example 2.13: Find the components along the coordinates of the position
vector of P( 4, 3)
Solution:
> > >
The position vector of P = OP =  4 i + 3 j
> >
Component of OP along xaxis is  4 i
i.e. component of OP along xaxis is a vector of magnitude 4 and its
direction is along the negative direction of xaxis.
> >
Component of OP along jaxis is 3 j
i.e. the component of OP along y axis is a vector of magnitude 3, having
its direction along the positive direction of yaxis.
> > >
Example 2.14: Express AB in terms of unit vectors i and j , where the
points are A( 6, 3) and B( 2,  5). Find also I AB I
Solution:
Given OA =  6 i + 3 j ; OB =  2 i 5j
:. AB = OB OA = {2~t5~f) (6~f + 3~f)
> >
= 4i 8y
IabI = U"f87l = V(4) 2 + ( 8) 2 =^16 + 64 =A /80
= 4^5
Theorem 2.11 (Without Proof) :
» > » >
If a , /? , c are three given noncoplanar vectors then every vector r in
» > > »
space can be uniquely expressed as r = Z a + m b + « c for some scalars Z,
m and n
61
Rectangular Resolution of a vector in three dimension
Theorem 2.12:
are unit vectors
If a point P in space has coordinate (x, y, z) then its position vector r is
>>> » Hi 5 2 >>>
xi + y y + z & and I r I = "\/;c + y + z where i , j , k
along OX, OY and OZ respectively.
Proof:
OX, OY, OZ are three mutually
» » »
perpendicular axes, i , y , & are unit
vectors along OX, OY, OZ respectively.
Let P be any point (x, y, z) in space and let
OP =~r
Draw PQ perpendicular to XOY
plane and QR perpendicular to OX
Then OR = x ; RQ = y ; QP = z
:. OR = x i ; RQ =yj; ^P = z k
U
V.
T
V
/
/
' ",'*'
s
''•.90°
A
Q
X
Fig. 2. 25
Now
OP = OQ + QP = OR + RQ + QP
— > » > >
OP = x j + y y + z &
> » > >
r = jc i +yy + z A:
»
Thus if P is a point (x, y, z) and r is the position vector of P, then
> » » >
r =ii + y _/' + z &
From the right angled triangle OQP, OP 2 =OQ 2 + QP 2
From the right angled triangle ORQ, OQ 2 =OR 2 + RQ 2
.. OP 2 = OR 2 + RQ 2 + QP 2
OP 2 = x 2 + y 2 + z 2
OP = \jx 2 + y 2 + z 2 => r = V?+
HI ./ 2^ 2^ 2
. . r = I r I = ^/x +y + z
2 , 2
y + z
2.6 Direction cosines and direction ratios
Let P(x, y, z) be any point in space with reference to a rectangular
coordinate system O (XYZ). Let a, (3 and y be the angles made by OP with the
positive direction of coordinate axes OX, OY, OZ respectively. Then cosa,
cos(3, cosy are called the direction cosines of OP .
62
In the fig 2.25 OQP = 90° ;  POZ = y :. \ OPQ = y (QPIIOZ)
PQ z x y
:. cosy = Qp => cosy = ~ Similarly cosa = ~ and cos(3 = ~;
> x y z /" ? 2 2
.". The direction cosines of OP are ~ ~, ~ where r = ^\Jx +y + z
Result 1: Sum of the squares of direction cosines is unity.
2 2o 2 (A 2 (jV fz\2 x 2 + y 2 + Z 2
cos^a + cos z p + cos z y = \jj + (£J + \^J = \
2
= % = 1 [v r 2 = x 2 + / + z 2 ]
r
2 2 2
.". cos a + cos (3 + cos y = 1
Result 2: Sum of the squares of direction sines is 2.
2 2 2 2 2 2
sin a + sin p + sin y = (1  cos a) + (1  cos p) + (1  cos y)
2 2 2
= 3  [cos a + cos P + cos y] = 31 =2
2 2 2
sin a + sin P + sin y = 2
Direction ratios:
Any three numbers proportional to direction cosines of a vector are called
its direction ratios, (d. r's).
> » > »
Let r =xi + y j + zk be any vector
 > x y z f 2 2 2
=> Direction cosines of r are ~ ~;, ~ where r = AJx + y + z
x y z
=> cos a = ~ ; cos p = ~. ; cos y = ~ where a, P, y be the angles made
»
by r with the coordinate axes OX, OY, OZ respectively
^ x y
osa i
_L_ __21
— r — — — r — r
cosa ' cosp ' cosy
cosa cosp cosy
=> x : y : z = cosa : cosp : cosy
i.e. the coefficients of i, j, k in the rectangular resolution of a vector are
proportional to the direction cosines of that vector.
> > > >
.'. x, y, z are the direction ratios of the vector r =x i +y I + z k
63
Addition, Subtraction and Multiplication of a vector by a scalar and
equality in terms of components:
> »—»» » > — » >
Let a = a j i + a 2 j + a^k and £> = b± i + b 2 j + 03 k be any two
vectors.
Then
> » > > >
(i) a + o = (fli + Pi) ' + ( a 2 + ^2)J + ( a 3 + 3) ' c
» » > .— » .>
(ii) a —b = (a^bAi +(^2~^2)J + ( a 3~^2) k
> f > > >
(iii) ma = m^a ^ i + a 2 j + a^k
> > »
= ma j i + ma 2 7 + »«*3 A; where m is a scalar
> >
(iv) a = b <=> a\ = b\, a 2 = b 2 and 03 = 03
Distance between two points:
Let A (jcj, yi, Zi) and B(x 2 , y 2 , z 2 ) be any two points
Then AB = OBOA
> > A f » > >
*2 * + ^2 7 + ^2 * y  \ x i i +y\j +zi k
» > >
= (x 2 x 1 ) 1 +(y 2 yi)j +(z 2 zi) k
:. The distance between A and B is AB = I AB
I ABl =
» > >
(*2  jc,) i + (y 2  y,) 7 + (z 2  zj) fc
= V(^2  *i) 2 + (yi  y\) 2 + (Z2  z i) 2
» — » — »
Example 2.15: Find the magnitude and direction cosines of 2 i 7 + 7 fc
Solution:
Magnitude of it ~J + l~t = \l~t  ~j + lt\ = aJ(2) 2 + ( l) 2 + (7) 2
= ^4+1+49 =^54 =3yf6
> > > 2 17
Direction cosines of 2 i 7 + 7 fc are t  7= '  t  7= ' t7=t
Example 2.16: Find the unit vector in the direction of 3 i + 4 j  12 k
64
Solution: Let a = 3 j + 4 7  12 k
\t\ = I3~t + 4~J  I2~k\= V(3) 2 + (4) 2 + ( 12) 2
= a/9 + 16 + 144 =Vl69 =13
» » » >
.,_,.. ,> . a a 3 i + 4 ./  12 k
Unit vector in the direction or a is a = ~ — ~ = f5
\?\
> > > » — » — »
Example 2.17: Find the sum of the vectors j 7 + 2 & and 2 i + 3j 4 k
and also find the modulus of the sum.
Solution:
> > > > > > > >
Let a = i — j +2k , b =2 i +3 j 4 k
> > /> > »\ / > > A > > >
a + £ =^j;'+2fe^+V,2j+3;4Jfe^=3j+2;'2A:
l~a+in = ^3 2 + 2 2 + ( 2) 2 = ^9 + 4 + 4
= y/l7
Example 2.18: If the position vectors of the two points A and B
are i +2 j 3 k and 2 i 4 j + k respectively then find I AB I
Solution:
If O be the origin, then OA = i + 2 7 3 k, OB =2 i  4 7 + k
AB = OB  OA
= (2~t4? + ~lt)  (~t + 2~f  3~k)
> > >
= i  6 7 + 4 jfe
IabI = V(l) 2 + ( 6) 2 + (4) 2 = a/53
Example 2.19: Find the unit vectors parallel to the vector  3 i + 4 7
Solution: Let a =  3 i + 4 j
\t\ = yj( 3) 2 + 4 2 = a/9 +16 =5
65
a a 1 > 1 ( > >
» a f 3 > 4 >
Unit vectors parallel to a are + a = ± \—r~ i +~F j
Example 2.20: Find the vectors of magnitude 5 units, which are parallel to the
vector 2i  j
> » >
Solution: Let a = 2 i  j
\7\=yJ2 2 + (l) 2 =^5
a a 1 / > A 2 » 1 >
> > A
Vectors of magnitude 5 parallel to 2 i  _/' = ± 5 a
» — » >
Example 2.21: Show that the points whose position vectors 2 i + 3 j 5k,
> > > > — » — »
3 i + _/' 2J and 6 i  5 _/' + 7 & are collinear.
Solution: Let the points be A, B and C and O be the origin. Then
> >>> — > >>> > > > — »
OA = 2 i + 3 y  5 jfe ; OB = 3 j + j  2 jfc ; OC = 6 i  5 j + 7 k
— > — > — > / > > A / > > A
.. AB = OB  OA = ^3 i + j  2 k )  \2 i + 3j 5 k )
» » >
= i 2j +3k
AC = OC  OA =\6i 5j +1 k) \2i + 3j 5 k )
AC = 4 j  8 j + 12 k = 4 V i  2 j + 3 k )
= 4 AB
Hence the vectors AB and AC are parallel. Further they have the
common point A.
.". The points A, B, C are collinear.
66
Example 2.22: If the position vectors of A and B are 3 i  7 j Ik and
>>>■ — >
5 z +4 j +3 k , find AB and determine its magnitude and direction cosines.
Solution:
Let O be the origin. Then
> » > » > > > »
OA = 3 j  7 y  7 Jfc , OB =5 i + 4j +3k
AB = OB  OA = ^5 i + 4 y + 3 k )  \3 i  7 j  7 k )
AB = 2 j + 11 j + 10 fc
I AB I = a/(2) 2 + (H) 2 + (10) 2 = 15
m ,. . • 2 11 10
The direction cosines are tt , tf , tt
EXERCISE 2.2
(1) Find the sum of the vectors 4 i +5j + k,2i + 4j  k and
> » »
3 i  4 y + 5 /c . Find also the magnitude of the sum.
» — » — » — » — » » — » — » » — » — » — »
(2) If a = 3 i  j 4 k, b =  2 i + 4 j  3 k and c = i + 2 j  k
find l2a  fc +3c I
(3) The position vectors of the vertices A, B, C of a triangle ABC are
respectively
> > » — » — » » » — » >
2 i + 3 7 + 4 k ,  i +2 j  k and 3 i 5 j +6 k
Find the vectors determined by the sides and calculate the length of the
sides.
(4) Show that the points whose position vectors given by
> > » > > > — » — »
(i)  2 i + 3 j +5 k , i +2j + 3 k , 1 i  k
» — » — » > » — » > >
(ii) z 2 j +3fe, 2 i +3y 4/c and 7 7 + 10 /c are collinear.
» — » — » > » — »
(5) If the vectors a =2i 3; and b = — 6 i +m; are collinear, find the
value of m.
> r >
(6) Find a unit vector in the direction of i + ^J3 j
67
> > » — » > » > > >
4 i + 5 7 + 6 fe , 5 i + 6 j + 4 k , 6 i + 4 j + 5 k . Prove that the
(7) Find the unit vectors parallel to the sum of 3 i  5 j + 8 k
> >
and  2 _/'  2 &
> » » » » » »
(8) Find the unit vectors parallel to 3a  2 fc + 4 c where a=3 i  j 4 k,
> » » » » » » »
&=2f+4y'3fe, c = i + 2 j  k
(9) The vertices of a triangle have position vectors
> > >
4i + 5 7 + 6 fc , 5
triangle is equilateral.
» » » » > > > > >
(10) Show that the vectors 2 i  j + k , 3 i  4 7  4 k , i 3j 5 k
form a right angled triangle.
(11) Prove that the points 2 i + 3 j +4k, 3 i +4j +2k, 4 i +2 j +3 k
form an equilateral triangle.
» » »
(12) If the vertices of a triangle have position vectors i +2y +3fc,
» » » » » — »
2 j + 3 j + k and 3 i + j + 2 k , find the position vector of its
centroid.
(13) If the position vectors of P and Q are i +3 j Ik
> > > >
and 5 i  2 7 + 4 k , find PQ and determine its direction cosines.
(14) Show that the following vectors are coplanar
> — » — » » — » > — » — »
(i) j  2 y + 3 k ,  2 i +3 j 4 k ,  j +2k
» — » » — » — » — » — » » — »
(ii) 5i + 6 j + 7 jfe , 7 i 87 +9k, 3 i +20 7 + 5 it
> » » » >
(15) Show that the points given by the vectors 4 i +5 7 + k,— j  k,
» » » » > »
3 j +9y + 4 & and  4 i + 4 7 + 4 & are coplanar.
> > > » » »
(16) Examine whether the vectors i +3 j + k , 2 i  j  k
> >
and 77 + 5 fc are coplanar.
68
3. ALGEBRA
3.1 Partial Fractions:
Definitions:
Rational Expression: An expression of the form ~r^ where p(x) and
q(x) * are polynomials in x is called a rational expression.
2
5x 2 3x +2x\
The expressions ~2 , ~~2 are examples for rational
x +3x + 2 x + x  22
expressions.
Proper Fraction: A proper fraction is one in which the degree of the
numerator is less than the degree of the denominator.
3 X+ 1 Ix + 9
The expressions 2 > ~~3 2 are examples for proper
x + 4x +3 x + x  5
fractions.
Improper Fraction: An improper fraction is a fraction in which the degree of
the numerator is greater than or equal to the degree of the denominator.
3 c 2 a 2
x + 5x + 4 x — x + 1
The expressions ~~ 2 , ~2 are examples for improper
x +2x + 3 x +x + 3
fractions.
Partial Fraction:
7
Consider the sum of ^ and
x — 2 x 1
We simplify it as follows:
7 5 l(x  1) + 5(x  2) 7x7 + 5x10 12* 17
jc2 + xl _ (jc2)(jc1) ~ (*2)(xl) ~(x2)(xl)
12x 17
Conversely the process of writing the given fraction , y\ (x  V\ as
7 5
2 + x _ i 1S known as splitting into partial fractions (or) expressing as
partial fractions.
A given proper fraction can be expressed as the sum of other simple
fractions corresponding to the factors of the denominator of the given proper
fraction. This process is called splitting into Partial Fractions. If the given
d(x\
fraction ~r^ is improper then convert into sum of a polynomial expression and
a proper rational fraction by dividing p(x) by q(x).
69
Working Rule :
Given the proper fraction ~r^ . Factorise q(x) into prime factors.
Type 1: Linear factors, none of which is repeated.
If a linear factor ax + b is a factor of the denominator q(x) then
corresponding to this factor associate a simple fraction , , where A is a
constant (A ^ 0).
i.e., When the factors of the denominator of the given fraction are all linear
factors none of which is repeated, we write the partial fraction as follows :
x + 3 A B
7 — , S \ I", — TIT = — TT + t — TT where A and B are constants to
(x + 5)(2x+l) x + 5 2x+\
be determined.
3x + 7
Example 3.1: Resolve into partial fractions j
x  3x + 2
2
The denominator x  3x + 2 can be factorised into linear factors.
2 2
x —3x + 2=x — x — 2x + 2 = x (x — 1)  2 (x  1) = (x  1) (x — 2)
3x + l A B
We assume 7 = r + T where A and B are constants.
jc 3jc + 2 x ~ l x ~ 2
3x + 7 A(x2) + B(xl)
^ x 2 3x + 2 " (xl)(*2)
^ 3x + 7 = A(x2) + B(xl) ...(1)
Equating the coefficients of like powers of x, we get
Coefficient of x : A + B = 3 ... (2)
Constant term :  2A  B = 7 ... (3)
Solving (2) and (3) we get
A = 10
B = 13
3x + 7 10 13 13 10
x
3x + 2 x — 1 x — 2 x — 2 x — 1
Note: The constants A and B can also be found by successively giving suitable
values for x.
To find A, put x = 1 in (1)
3(1) + 7 = A(l2) + B(0)
10 = A(l)
A = 10
70
To find B, put x = 2 in (1)
3(2) + 7 = A(0) + B(2  1)
B = 13
3x + 7 10 13
x l  3x + 2 ~xl x2
3x + 7 13 10
~7T
x + 4
3x + 2 ^  2 x  1
Example: 3.2: Resolve into partial fractions — r
(x  4) (x + 1)
2
The denominator (x  4) (x + 1) can be further factored into linear factors
i.e. (x 2  4) (x + 1) = (x + 2) (x 2)(x+ 1)
x + 4 ABC,
Let — 7 = — ~z + + — — r, where A, B and C are
(x 4)(x+l) x + 2 x2 x+l'
constants to be determined.
x + 4 _ A(x  2) (x + 1) + B(x + 2) (x + 1) + C(x + 2) (x  2)
(x 2 4)(x+l) ~ (x + 2) (x  2) (x + 1)
=> x + 4 = A(x2)(x+l) + B(x + 2)(x+l) + C(x + 2)(x2) ...(1)
To find A, put x =  2 in (1)
2 + 4 = A(22)(2+l) + B(0) + C(0)
2 = 4A => A = 1/2
To find B, put x = 2 in (1), we get B = 1/2
To find C, put x =  1 in (1), we get C =  1
• x + 4 _ 1/2
(x 2 4)(x+l) ~ ( X+T >
_ x + 4 __ 1
^ (x 2 4)(x+l) " 2 ^ + 2 ) + 2(x2) "x+l
Type 2: Linear factors, some of which are repeated
If a linear factor ax + b occurs n times as a factor of the denominator of the
given fraction, then corresponding to these factors associate the sum of n simple
fractions,
Ai A 2 A3 A n
1,
'Z ( 1;
(x
2) """x+l
1 1
+ + o +...+'
ax + b (ax + b) 2 (ax + bf ■'■ (ax + bf
Where A1 , A2, A3, ... A„ are constants.
71
9
xumple 3.3: Resolve into partial fi actions ')
(x  1) (x + 2)
9 ABC
Lei •> — , + , i + ')
(xl)(x + 2) x ~ l x + 1 (x + 2)
9 A(x + 2) 2 + B(x  1) (x + 2) + C(x 
1)
(x  1) (x + 2) 2 ~ (x  1) (x + 2) 2
=> 9 = A(x + 2) 2 + B(x  1) (x + 2) + C(x 
1)
To find A, put x = 1 in (1)
We get 9 = A (1 + 2) 2 => A = 1
To find C, put x = 2 in (1)
We get 9 = C ( 2  1) => C =  3
In (1), equating the coefficient of x on both sides we get
A + B =
=> 1+B = =>B = 1
9 113
" (x  1) (x + 2) Z ~ x  1 x + 2 ( x + 2) 2
(1)
Type 3: Quadratic factors, none of which is repeated
2
If a quadratic factor ax + bx + c which is not factorable into linear factors
occurs only once as a factor of the denominator of the given fraction, then
Ax+B
corresponding to this factor associate a partial fraction — 2~^ where A
ax + bx + c
and B are constants which are not both zeros.
Consider 1
(x + 1) (x + 1)
• , • £■■,* 2x A Bx + C
We can write this proper fraction in the form 2 = I i +~ 2
(x+ 1) (x + 1) x+ 1 x+1
The first factor of the denominator x + 1 is of first degree, so we assume its
numerator as a constant A. The second factor of the denominator x +1 is of 2
degree and which is not factorable into linear factors. We assume its numerator
as a general firstdegree expression Bx + C.
x  2x  9
Example 3.4: Resolve into partial fractions — 2
(x + x + 6) (x + 1)
72
2
x 2x9 Ax + B C
Let —2 = ~2 + v 7 i
(x +x + 6)(x+l) x +x + 6 x+1
x 2 2x9 (Ax + B)(x+ 1) + C(x 2 + x + 6)
=> ~ 2 = 2
(x +x + 6) (x+ 1) (x +x + 6) (x + 1)
=> x 2 2x9 = (Ax+B)(x+1) +C(x 2 + x + 6) ... (1)
To find C put x = — 1 in (1)
We get l+29 = C(ll+6)=>C=l
To find B, put x = in (1)
We get  9 = B + 6C
9 = B6 => B = 3
To find A, Put jc = 1 in (1)
1  2  9 = (A  3) (2) + ( 1) (8) =>  10 = 2A  14
A = 2
* 2x9 2x  3 1
~2 ;tt — r: =t
(x + x + 6)(x+l) x +x + 6 x+ ^
2
x + x + 1
Example 3.5: Resolve into partial fractions ~2
x  5x + 6
Solution:
Here the degree of the numerator is same as the degree of the denominator,
i.e. an improper fraction.
x
On division ~2 — ~ 7 = 1 + 2" \ •■■(!)
Let
x 
6x  5 = A(x  3) + B(x  2)
By putting x = 2, A=125 =^> A =  7
By putting x = 3, B = 185 =^ B = 13
2
x +x + 1
6x5
 1 +T
x  5x + 6
A B
x  2 x  3
x  5x + 6
6x 5
r — rijc + n
• (1)
2
X
+ X+ 1
=
1
7
x 
X
2
7
13
2
x 
2
X
5x + 6
+ x+ 1
X 3
13
2
x ■
5x + 6
2 ~^x3
73
7xl
(i)
6  5x + x
x2
(x + 2) (jc  l) 2
2
2x  5x — 7
(x2) 3
2
7x  25* +
6
2
x + x + 1
(JC
(6)
(9)
■ (12;
1)(JC
2)(x
x+ 1
■3)
(X
 2) 2 (x +
x 2 3
3)
(x + 2) (x +
2
X +x + 1
1)
• 2
EXERCISE 3.1
Resolve into partial fractions
(1) (xlAx+l) ™
(4) : n ) m1 .i (5)
(x 1) (x + 2)
2
x  6x + 2
(7) ^TT^T (8)
x (x + 2)
x + 2
(10) 2 (11)— 2"
(x + 1) (x + 1) (x  2x  1) (3x  2) x + 2x +1
3.2 Permutations:
Factorial:
The continued product of first w natural numbers is called the
"n factorial" and is denoted by n ! or \n_
i.e. n\ = 1x2x3x4x...x(m1)xm
5! = 1x2x3x4x5= 120
Zero Factorial:
We will require zero factorial in the latter sections of this chapter. It does
not make any sense to define it as the product of the integers from 1 to zero. So,
we define ! = 1 .
Deduction:
n\ = Ix2x3x4x...x(wl)xw
= [Ix2x3x4x...x(n l)]n
= [(«l)!]«
Thus, n\ = n [(n  1)!]
For example,
8 ! = 8(7 !)
3.2.1 Fundamental Principles of Counting:
In this section we shall discuss two fundamental principles viz., principle
of addition and principle of multiplication. These two principles will enable us
to understand permutations and combinations and form the base for
permutations and combinations.
74
Fundamental Principle of Multiplication:
If there are two jobs such that one of them can be completed in m ways,
and when it has been completed in any one of these m ways, second job can be
completed in n ways; then the two jobs in succession can be completed in
m x n ways.
Explanation:
If the first job is performed in any one of the m ways, we can associate
with this any one of the n ways of performing the second job: and thus there are
n ways of performing the two jobs without considering more than one way of
performing the first; and so corresponding to each of the m ways of performing
the first job, we have n ways of performing the second job. Hence, the number
of ways in which the two jobs can be performed is m x n.
Example 3.6: In a class there are 15 boys and 20 girls. The teacher wants to
select a boy and a girl to represent the class in a function. In how many ways
can the teacher make this selection?
Solution :
Here the teacher is to perform two jobs :
(i) Selecting a boy among 15 boys, and
(ii) Selecting a girl among 20 girls
The first of these can be performed in 15 ways and the second in 20 ways.
Therefore by the fundamental principle of multiplication, the required
number of ways is 15 x 20 = 300.
Fundamental Principle of Addition:
If there are two jobs such that they can be performed independently in
m and n ways respectively, then either of the two jobs can be performed in
(m + n) ways.
Example 3.7: In a class there are 20 boys and 10 girls. The teacher wants to
select either a boy or a girl to represent the class in a function. In how many
ways can the teacher make this selection?
Solution:
Here the teacher is to perform either of the following two jobs :
(i) selecting a boy among 20 boys, (or)
(ii) Selecting a girl among 10 girls
The first of these can be performed in 20 ways and the second in 10 ways.
Therefore, by fundamental principle of addition either of the two jobs can
be performed in (20 + 10) = 30 ways.
75
Hence, the teacher can make the selection of either a boy or a girl in 30
ways.
Example 3.8: A room has 10 doors. In how many ways can a man enter the
room through one door and come out through a different door?
Solution:
Clearly, a person can enter the room through any one of the ten doors. So,
there are ten ways of entering into the room.
After entering into the room, the man can come out through any one of
the remaining 9 doors. So, he can come out through a different door in 9 ways.
Hence, the number of ways in which a man can enter a room through one
door and come out through a different door = 10 x 9 = 90.
Example 3.9: How many words (with or without meaning) of three distinct
letters of the English alphabets are there?
Solution:
Here we have to fill up three places by distinct letters of the English
alphabets. Since there are 26 letters of the English alphabet, the first place can
be filled by any of these letters. So, there are 26 ways of filling up the first
place.
Now, the second place can be filled up by any of the remaining 25 letters.
So, there are 25 ways of filling up the second place.
After filling up the first two places only 24 letters are left to fill up the
third place. So, the third place can be filled in 24 ways.
Hence, the required number of words
= 26 x 25 x 24 = 15600
Example 3.10:
How many threedigit numbers can be formed by using the digits 1, 2, 3, 4, 5.
Solution :
We have to determine the total number of three digit numbers formed by
using the digits 1, 2, 3, 4, 5.
Clearly, the repetition of digits is allowed.
A three digit number has three places viz. unit's, ten's and hundred's. Unit's
place can be filled by any of the digits 1, 2, 3, 4, 5. So unit's place can be filled
in 5 ways.
Similarly, each one of the ten's and hundred's place can be filled in 5 ways.
.•. Total number of required numbers
= 5 x 5 x 5 = 125
76
Example 3.11: There are 6 multiple choice questions in an examination. How
many sequences of answers are possible, if the first three questions have
4 choices each and the next three have 5 each?
Solution:
Here we have to perform 6 jobs of answering 6 multiple choice questions.
Each of the first three questions can be answered in 4 ways and each one of
the next three can be answered in 5 different ways.
So, the total number of different sequences
= 4x4x4x5x5x5 = 8000
Example 3.12: How many threedigit numbers greater than 600 can be formed
by using the digits 4, 5, 6, 7, 8?
Solution:
Clearly, repetition of digits is allowed. Since a threedigit number greater
than 600 will have 6, 7 or 8 at hundred's place. So, hundred's place can be
filled in 3 ways.
Each of the ten's and one's place can be filled in 5 ways.
Hence, total number of required numbers
= 3x5x5 = 75
Example 3.13: How many numbers divisible by 5 and lying between 5000 and
6000 can be formed from the digits 5, 6, 7, 8 and 9?
Solution:
Clearly, a number between 5000 and 6000 must have 5 at thousand's place.
Since the number is divisible by 5 it must have 5 at unit' s place.
Now, each of the remaining places (viz. Hundred's and ten's) can be filled in
5 ways.
Hence the total number of required numbers
= 1x5x5x1= 25
Example 3.14: How many three digit odd numbers can be formed by using the
digits 4, 5, 6, 7, 8, 9 if :
(i) the repetition of digits is not allowed?
(ii) the repetition of digits is allowed?
Solution:
For a number to be odd, we must have 5, 7 or 9 at the unit's place.
So, there are 3 ways of filling the unit's place.
(i) Since the repetition of digits is not allowed, the ten's place can be filled
with any of the remaining 5 digits in 5 ways.
Now, four digits are left. So, hundred's place can be filled in 4 ways.
77
So, required number of numbers
= 3x5x4 = 60
(ii) Since the repetition of digits is allowed, so each of the ten's and
hundred's place can be filled in 6 ways.
Hence required number of numbers = 3x6x6=1 08
EXERCISE 3.2
1. In a class there are 27 boys and 14 girls. The teacher wants to select
1 boy and 1 girl to represent a competition. In how many ways can
the teacher make this selection?
2. Given 7 flags of different colours, how many different signals can be
generated if a signal requires the use of two flags, one below the
other?
3. A person wants to buy one fountain pen, one ball pen and one pencil
from a stationery shop. If there are 10 fountain pen varieties, 12 ball
pen varieties and 5 pencil varieties, in how many ways can he select
these articles?
4. Twelve students compete in a race. In how many ways first three
prizes be given?
5. From among the 36 teachers in a college, one principal, one vice
principal and the teacherin charge are to be appointed. In how many
ways this can be done?
6. There are 6 multiple choice questions in an examination. How many
sequences of answers are possible, if the first three questions have 4
choices each and the next three have 2 each?
7. How many numbers are there between 500 and 1000 which have
exactly one of their digits as 8?
8. How many fivedigit number license plates can be made if
(i) first digit cannot be zero and the repetition of digits is not
allowed.
(ii) the first digit cannot be zero, but the repetition of digits is
allowed?
9. How many different numbers of six digits can be formed from the
digits 2, 3, 0, 7, 9, 5 when repetition of digits is not allowed?
10. How many odd numbers less than 1000 can be formed by using the
digits 0, 3, 5, 7 when repetition of digits is not allowed?
11. In how many ways can an examinee answer a set of 5 true / false
type questions?
78
12. How many 4digit numbers are there?
13. How many three  letter words can be formed using a, b, c, d, e if :
(i) repetition is allowed (ii) repetition is not allowed?
14. A coin is tossed five times and outcomes are recorded. How many
possible outcomes are there?
3.2.2. Concept of Permutations:
The word permutation means arrangement.
For example, given 3 letters a, b, c suppose we arrange them taking 2 at a
time.
The various arrangements are ab, ba, be, cb, ac, ca.
Hence the number of arrangements of 3 things taken 2 at a time is 6 and this
can be written as 3P2 = 6.
Definition:
The number of arrangements that can be made out of n things taking r at a
time is called the number of permutations of n things taken r at a time.
Notation:
If n and r are positive integers such that 1< r < n, then the number of all
permutations of n distinct things, taken r at a time is denoted by the symbol P(w,
r) or nPr.
We use the symbol nVr throughout our discussion. Thus nPr = Total number
of permutations of n distinct things taken r at a time.
Note: In permutations the order of arrangement is taken into account; when the
order is changed, a different permutation is obtained.
Example 3.15: Write down all the permutations of the vowels A, E, I, O, U in
English alphabets taking 3 at a time and starting with E.
Solution: The permutations of vowels A, E, I, O, U taking three at a time and
starting with E are
EAI, EIA, EIO, EOI, EOU, EUO, EAO, EOA, EIU, EUI, EAU, EUA.
Clearly there are 12 permutations.
Theorem 3.1:
Let r and n be positive integers such that 1 < r < n.
Then the number of all permutations of n distinct things taken r at a time is
given by n(n  1) (n — 2). . . \n  r l)
i.e. nPr = n(n — 1) (n — 2) ... (n — r l)
79
Proof:
Let nPr denote the number of permutations of n things taken r at a time.
Clearly the total number of permutations required is same as the number of
possible ways of filling up r blank spaces by n things.
□ □□•■•□
12 3 r
Let there be r blank spaces arranged in a row
The first place can be filled by any one of the n things in n ways.
If the first place is filled up by any one of the n things, there will be
(n  1) things remaining. Now the second place can be filled up by any one of
the (n  1) remaining things.
Here it can be filled up in (n  1) ways.
Hence the first two places can be together filled in n(n  1) ways.
Having filled up these two places, we have (n  2) things remaining with
which we can fill up the third place. So the third place can be filled up by any
one of these things in (n  2) ways.
Hence the first three places can be together filled in n(n  1) (n — 2) ways.
Proceeding in this way, we find that the total number of ways of filling up
the r spaces is
n(n  1) (n — 2). . . upto r factors
i.e. n(n  I) (n2) ... \n — r—l)
.". nPr = n(n  Y) (n2) ... \n r \) = n(n  1) (n — 2) ... (n r + 1)
Theorem 3.2:
n\
Let r and n be positive integers such that 1 < r < n. Then nVr = —
r ° (nr)\
Proof:
riPr = n(n  I) (n2) ... (n r l)
n(n  1) (n  2)... (n  rl) (n r)(n r"+l)...2.1
(nr) (nr+l) ... 2.1
n\
~ (nr)\
80
Theorem 3.3:
The number of all permutations of n distinct things, taken all at a time is n\
Proof: nPr = n(n 1) (n2) ... \n rl)
By putting r = n, nPn = n(n  I) (n2) ... (n n l)
= n(n  1) (n  2) ...(« n l)
= n(nl)(n2) ... 1
= n\
:. riPn = n\
Remark: We have already defined 0! = 1. This can also be derived as follows.
We know that
By putting r = n,
P n]
nT '(nr)\
nVn  , ,,
(n — n)\
=>
n\
n\ = Tjj (v nPn = n\)
=>
n\
0! = ^y = 1
0! = 1
Example 3.16: Evaluate
8P 3
Solution:
8! 8! (8x7x6)x5!
aVi "(8 3)! _ 5!  5!
=8x7x6
= 336
Example 3.17 : If 5Pr =6P r _i, find r
Solution: 5Pr = 6P r _i
5! 6!
(5r)! ( 6 _i),
5! 6x5!
(5r)!  (7r)!
5! 6x5!
(5r)! {(7r)(6r)}(5r)!
1= 6 
(7r)(6r)
=> (7r)(6r) = 6 => 42 7r6r+ ?6 =
=> r 2 13r+36 = => (r9)(r4) =
=> r=9orr=4
=> r = 4 ( v 5Pr is meaningful for r < 5)
Example 3.18:
If WP4 = 360, find the value of n.
n !
Solution: uPa = 360 => ; 777 =6x5x4x3
(m  4) !
n ! 6x5x4x3x2! 6!
" (n4)! " 2!  2!
=> n! = 6!
^> « = 6
Example 3.19:
If 9Pr = 3024, find r.
Solution: 9Pr = 3024
=^> =9x8x7x6 = 9P 4
^> r = 4
Example 3.20:
If(nl)P3 :«P4 = 1 :9, find «.
Solution:
(n  1)P3 : MP4 = 1 : 9
=> (n  1) (n  2) («  3) : n(n  1) (n  2) (n  3) = 1 : 9
=> i.e. 9(m  1) (n  2) (n  3) = «(«  1) (n  2) (n  3)
^« = 9
Example 3.21: In how many ways can five children stand in a queue?
Solution:
The number of ways in which 5 persons can stand in a queue is same as the
number of arrangements of 5 different things taken all at a time.
Hence the required number of ways
= 5 P5 = 5! =120
Example 3.22: How many different signals can be made by hoisting 6
differently coloured flags one above the other, when any number of them may
be hoisted at one time?
82
Solution:
The signals can be made by using at a time one or two or three or four or
five or six flags.
The total number of signals when rflags are used at a time from 6 flags is
equal to the number of arrangements of 6, taking r at a time i.e. 6P r
Hence, by the fundamental principle of addition, the total number of
different signals
= 6Pl + 6P2 + 6P3 + 6P4 + 6P5 + 6P6
= 6 + (6x5) + (6x5x4) + (6 x5x4x 3) + (6x5x4x3 x2)
+ (6x5x4x3x2x1)
= 6 + 30 + 120 + 360 + 720 + 720 = 1956
Example 3.23: Find the number of different 4letter words with or without
meanings, that can be formed from the letters of the word 'NUMBER'
Solution:
There are 6 letters in the word 'NUMBER' .
So, the number of 4letter words
= the number of arrangements of 6 letters taken 4 at a time
= 6P4
= 360
Example 3.24: A family of 4 brothers and 3 sisters is to be arranged in a row,
for a photograph. In how many ways can they be seated, if
(i) all the sisters sit together.
(ii) all the sisters are not together.
Solution :
(i) Since the 3 sisters are inseparable, consider them as one single unit.
This together with the 4 brothers make 5 persons who can be arranged
among themselves in 5! ways.
In everyone of these permutations, the 3 sisters can be rearranged among
themselves in 3! ways.
Hence the total number of arrangements required = 5! x 3! = 120 x 6 = 720
(ii) The number of arrangements of all the 7 persons without any restriction
=7! =5040
Number of arrangements in which all the sisters sit together = 720
.". Number of arrangements required = 5040  720 = 4320
83
3.2.3 Permutations of objects not all distinct:
The number of mutually distinguishable permutations of n things, taken all
at a time, of which p are alike of one kind, q alike of second such that p + q = n,
. n\
is — ; — r
p\q\
Example 3.25: How many permutations of the letters of the word 'APPLE' are
there?
Solution:
Here there are 5 letters, two of which are of the same kind.
The others are each of its own kind.
5! 5! 120
.". Required number of permutations is = tmTTTTT = ?7 = ~ 1 ?~ = ^
Example 3.26: How many numbers can be formed with the digits 1, 2, 3, 4, 3,
2, 1 so that the odd digits always occupy the odd places?
Solution:
There are 4 odd digits 1, 1, 3, 3 and 4 odd places.
4!
So odd digits can be arranged in odd places in ~, ~, ways.
3!
The remaining 3 even digits 2, 2, 4 can be arranged in 3 even places in ~7
ways.
4! 3!
Hence, the required number of numbers = 9T2T X V =6x3 = 18
Example 3.27: How many arrangements can be made with the letters of the
word "MATHEMATICS"?
Solution:
There are 11 letters in the word 'MATHEMATICS' of which two are M's,
two are A's, two are T's and all other are distinct.
11!
.•. required number of arrangements = — — — — — = 4989600
3.2.4 Permutations when objects can repeat:
The number of permutations of n different things, taken r at a time, when
f
each may be repeated any number of times in each arrangement, is n
Consider the following example:
In how many ways can 2 different balls be distributed among 3 boxes?
Let A and B be the 2 balls. The different ways are
84
ox 1
Box 2
Box 3
A
B
D
m
A
D
n
A
U
□
m
A
s
n
B
s
n
A
D
D
AB
AB
D
D
□
AB
D
i.e. 9 ways. By formula n r = 3 =9 ways
Example 3.28: In how many ways can 5 different balls be distributed among
3 boxes?
Solution:
There are 5 balls and each ball can be placed in 3 ways.
So the total number of ways = 3 = 243
Example: 3.29: In how many ways can 3 prizes be distributed among 4 boys,
when (i) no boy gets more than one prize?
(ii) a boy may get any number of prizes?
(iii) no boy gets all the prizes?
Solution:
(i) The total number of ways is the number of arrangements of 4 taken 3 at
a time.
So, the required number of ways = 4P3 = 4 ! = 24
(ii) The first prize can be given away in 4 ways as it may be given to
anyone of the 4 boys.
The second prize can also be given away in 4 ways, since it may be obtained
by the boy who has already received a prize.
Similarly, third prize can be given away in 4 ways.
Hence, the number of ways in which all the prizes can be given away
= 4x4x4 =4 3 =64
85
(iii) Since any one of the 4 boys may get all the prizes. So, the number of
ways in which a boy get all the 3 prizes = 4.
So, the number of ways in which a boy does not get all the prizes = 644=60
3.2.5 Circular Permutations:
We have seen that the number of permutations of n different things taken all
together is n\, where each permutation is a different arrangement of the n things
in a row, or a straight line. These permutations are called linear permutations or
simply permutations.
A circular permutation is one in which the things are arranged along a circle.
It is also called closed permutation.
Theorem 3.4:
The number of circular permutations of n distinct objects is (n 1)!
Proof:
Letai, Q2, •■• , cinh a n be n distinct objects.
Let the total number of circular permutations be x.
Consider one of these x permutations as shown in figure.
Clearly this circular permutation provides n
near permutations as given below
a\, ci2, ai, ... , a n  \, a n
Cl2, (23, «4,
«3, «4, a$,
a n ,a\, C12,
a n >
a 2
an I
Fig. 3. 1
Thus, each circular permutation gives n linear permutations.
But there are x circular permutations.
So, total number of linear permutations is xn.
But the number of linear permutations of n distinct objects is n\.
:. xn = n\
x =
nl_
n
x = (n  1) !
.". The total number of circular permutations of n distinct objects is (n  1)!
Note: In the above theorem anticlockwise and clockwise order of arrangements
are considered as distinct permutations.
86
Difference between clockwise and anticlockwise arrangements:
Consider the following circular permutations:
Fig. 3. 2 Fig. 3. 3
We observe that in both, the order of the circular arrangement is a u a 2 , a 3 , a 4 .
In fig (3.2) the order is anticlockwise, whereas in fig. (3.3) the order is
clockwise.
Thus the number of circular permutation of n things in which clockwise and
anticlockwise arrangements give rise to different permutations is (n  1)!
If there are n things and if the direction is not taken into consideration, the
number of circular permutations is ^ (/i — 1)!
Example 3.30:
In how many ways 10 persons may be arranged in a (i) line (ii) circle?
Solution:
(i) The number of ways in which 10 persons can be arranged in a line
= loPlO = 10!
(ii) The number of ways in which 10 persons can be arranged in a circle
= (101)!= 9!
Example 3.31: In how many ways can 7 identical beads be stung into a ring?
Solution: Since the arrangement is circular either clockwise arrangement or
anticlockwise arrangement may be considered.
1 6!
.". The required number of ways = ~ (7 1)! ="9" = 360
Example 3.32: In how many ways can 5 gentlemen and 5 ladies sit together at a
round table, so that no two ladies may be together?
Solution:
The number of ways in which 5 gentlemen may be arranged is (5  1)! = 4!
87
Then the ladies may be arranged among themselves in 5! ways.
Thus the total number of ways =4!x5!=24xl20 = 2880
Example 3.33: Find the number of ways in which 8 different flowers can be
strung to form a garland so that 4 particular flowers are never separated.
Solution:
Considering 4 particular flowers as one flower, we have five flowers, which
can be strung to form a garland in 4 ! ways.
But 4 particular flowers can be arranged in 4! ways.
.". Required number of ways = 4 ! x 4 ! = 576
EXERCISE 3.3
1 . Evaluate the following :
(i) 5 P 3 (ii) 15P3 (iii) 5P5 (iv)25P20 (v) 9P5
2. If W P 4 = 20 . W P3 , findw.
3. If ioP r =5040, find the value of r.
4. If 56P(r + 6) :54P(r + 3) = 30800 : 1, find r
5. If P m stands for m P m , then prove that 1 + 1 .Pi + 2.P2 + 3.P3 + ...
+ n.P n = («+ 1)!
6. Prove that n V r = (n _ i)P r + r . („ _ i)P( r _ i).
7. Three men have 4 coats, 5 waistcoats and 6 caps. In how many ways can
they wear them?
8. How many 4letter words, with or without meaning, can be formed, out
of the letters of the word, 'LOGARITHMS', if repetition of letters is not
allowed?
9. How many 3digit numbers are there, with distinct digits, with each digit
odd?
10. Find the sum of all the numbers that can be formed with the digits
2, 3, 4, 5 taken all at a time.
11. How many different words can be formed with the letters of the word
'MISSISSIPPI'?
12. (i) How many different words can be formed with letters of the word
'HARYANA'?
(ii) How many of these begin with H and end with N?
13. How many 4digit numbers are there, when a digit may be repeated any
number of times?
14. In how many ways 5 rings of different types can be worn in 4 fingers?
15. In how many ways can 8 students are seated in a (i) line (ii) circle?
16. In how many ways can a garland of 20 similar flowers are made?
3.3 Combinations:
The word combination means selection. Suppose we are asked to make a
selection of any two things from three things a, b and c, the different selections
are ab, be, ac.
Here there is no reference to the order in which they are selected.
i.e. ab and ba denote the same selection. These selections are called
combinations.
Definition:
A selection of any r things out of n things is called a combination of
n things r at a time.
Notation:
The number of all combinations of n objects, taken r at a time is generally
denoted by n C r or C(n,r) or (1 . We use the symbol n C r throughout our
discussion.
(Number of ways of selecting
Thus n C r = j r ob j ects from „ bj e cts
Difference between Permutation and Combination:
1. In a combination only selection is made whereas in a permutation not
only a selection is made but also an arrangement in a definite order is
considered.
i.e. in a combination, the ordering of the selected objects is immaterial
whereas in a permutation, the ordering is essential.
2. Usually the number of permutation exceeds the number of combinations.
3. Each combination corresponds to many permutations.
Combinations of n different things taken r at a time:
Theorem 3.5:
The number of all combinations of n distinct objects, taken r at a time is
n\
given by n C r = ~ — — r
b J n (nr) ! r!
Proof: Let the number of combinations of n distinct objects, taken r at a time be
denoted by n C r .
Each of these combinations contains r things and all these things are
permuted among themselves.
.". The number of permutations obtained is r !
Hence from all the n C r combinations we get n C r x r! permutations.
But this gives all the permutations of n things taken r at a time i.e. n P r .
Hence, n C r . r ! = „P r
■ • nW —
«P.
r
r
"'■ < r, M '
i.Pr =
(nr)\r\ \' n r (nr)\
Properties
(1)„C„=1 (2)„Co=l (3) n C r = n C n  r 0<r<n
Proof:
n\
(1) We know that n C r =
(n  r)\ r\
Putting r = n, we have „C W = (n _ n)!n! = q!^!
= 1
(2) Putting r = 0, we have
n\ nj
(n0) !0! = «!
m! m!
"^O  /„_mim  „!  1
(3) We have „C W _ r = N  ...
/ \ («  r)! r!
(n — r)\ \n— n — r )\
= rrr
Note: The above property can be restated as follows :
If x and y are nonnegative integers such that x + y = n, then n C x = n Cy
(4) If n and r are positive integers such that r < n,
then n C r + n C( r  l) = («+l)C r
Proof: We have
rr~r + rp~(r— 1) —
(nr)!r! („ _ —{), (r _ iy
{nr)\r\ (nr+l) !(rl)
+
(nr) !r{(rl)!} (n  r + 1) {(n  r)\ (r 1)!
90
(1 1
(wr)!(rl)! [r nr+1
n\ In — r + 1 + r
(nr)\(rl)\ [r(nr+l)
n\ \ n + 1
(nr)\{r\)\ [r(n  r + I)
(n+l) \n\]
~ (n  r + 1) (n  r)\ r(r  1)!
(m+1)!
~~ (n r+ 1)! r!
(ra+1)!
~~ (« + 1  r) ! r!
= in + l)C r
(5) If n and r are positive integers such that 1 < r < n,
then n C r = ~ (nl)C(rl)
Proof:
C  " !
npi1)!
[(nl)(rl)]!r(rl)l
n (»!)!
r [(„l)(rl)]!(rl)]
r (nl)C(rl)
(6)If 1 < r < n, then « . („ _ i)C( r  1) = (n  r + 1) . „C( r _ i)
Proof:
We have n. (n . 1)C(r _ 1} = „ {^ _  ^^  _ 
77
(nr)!(rl)!
(n r + \)n\
(nr+l)(nr)!(rl)!
91
;/
(nr+l)\ (r1)!.
= (nr+Y)
= ( " r+1) >rl)!(rl)!
= (nr+ 1) . w C( r _ i)
(7) For any positive integers x and y,
n C x = n C y ^ x = y or x + y = n
Proof: We have n C x = n C y
— > nSx = rpy = nS'in  y) V '■' rry = n^(n  y)\
=^>x = y or x = ny
=>x = y or x + y = n
Note: If n C x = nCy and x ^ y, then x + y = n
Example 3.34: Evaluate the following :
5
(ii) I 5C r
(i) 6C 3
Solution:
0)
(ii)
r=\
6C3 =
6P3 6x5x4
3!
1x2x3
= 20
Z 5C r = 5C1 + 5C2 + 5C3 + 5C4 + 5C5
r=l
= 5 + 10+10 + 5+1=31
Example 3.35: If W C4 = n C(, , find i2C n
Solution:
W C4 = n C(, => « = 4 + 6 = 10
Now i 2 C„ = 12C10
12x11
= 12C(12  10) = 12C2 = i x 2
= 66
Example 3.36: If 15CV : isC( r _ i) = 11:5, find r
92
Solution:
„  15Q 11
rljxx.,  15 C (r _i) 
15!
5
r!(15r)!
11
15!
5
(rl)!(15r+ 1)!
15! (rl)!(16r)!
11
r!(15r)! x 15!
(rl)!(16r){(15r)!}
5
11
r(r 1)! (15  r)!
16 r
r
5
11
5
=> 5(16 r) = llr => 80 = 16r
=> r= 5
Example 3.37:Show that the product of r consecutive integers is divisible by r!
Solution:
Let the r consecutive integers be « + l,n + 2,n + 3, ...,n + r
Hence their product = (n+ 1) (n + 2) (n + 3) ... (n + r)
_ 1.2.3. .. n.(n+ l)(w + 2) ... (w + r)
1.2.3... n
_ (n + r) !
= re!
their product _ (n + r) !
r! ~ n\r\
= (n + r )C r which is an integer.
.". The product of r consecutive integers is divisible by r!
Example 3.38: Let r and n be positive integers such that 1 < r < n. Then prove
the following :
«C r n — r + i
77
C(rl) r
«C r r!(nr)!
Solution:
n
C(rl)
(rl)!(nr+l)!
93
n\ (rl)!(wr+l)!
~ r\(nr)\ * n\
(rl)!Qir+l) {(nr)!}
r(rl)!(wr)!
n — r + 1
r
Example 3.39 : If n P r = n P( r + i) and n C r = w C( r  1) , find the values of n and r
Solution:
n!
(1)
w P r  n P(r+ 1) =>
(nr)!
1
~ fnr1)!
1
=>
(«
r)(wrl)!
~ (nrl)\
=>
rrr — n*(r — 1) — ^
«  r
n!
= 1
... i
r! (n  r) !
~ (rl)\(nr+l)\
«!
n\
^>
r(rl)!(wr)!
~ (rl)!(nr+l) {(«
r)!}
1
1
r ~ nr+l
=> n  r + 1 = r
=> w2r =1 ...(2)
Solving (1) and (2) we get n = 3 and r = 2
EXERCISE 3.4
1. Evaluate the following:
(i) 10C8 (ii) 100C98 (iii) 75C75
2. If n Cio = w Ci2, find23C n
3. If sCr7C3 = 7C2, find r
4. If i 6 C 4 =i 6 C r + 2,find r C2
20
5. Find n if (1) 2 . n C?, = y „C2 (11) „C(„ _ 4) =70
6. If („ + 2)Cs : (n  2)?4 = 57 : 16, find n.
7. If 28C2/ : 24C(2r  4) = 225 : 11, find r.
94
Practical problems on Combinations
Example 3.40: From a group of 15 cricket players, a team of 11 players is to be
chosen. In how many ways this can be done?
Solution:
There are 15 players in a group. We have to select 11 players from the
group.
.". The required number of ways = 15C1 1
15xl4xl3x 12 ,„„
15Cl1 = 1x2x3x4 = 1365wa y s
Example 3.41: How many different teams of 8, consisting of 5 boys and 3 girls
can be made from 25 boys and 10 girls?
Solution:
5 boys out of 25 boys can be selected in 25C5 ways.
3 girls out of 10 girls can be selected in 10C3 ways.
.". The required number of teams = 25C5 x 10C3 = 6375600
Example 3.42: How many triangles can be formed by joining the vertices of a
hexagon?
Solution:
There are 6 vertices of a hexagon.
One triangle is formed by selecting a group of 3 vertices from given
6 vertices.
This can be done in 6C3 ways.
6!
.•. Number of triangles = 6C3 = 3T37 = 20
Example 3.43:
A class contains 12 boys and 10 girls. From the class 10 students are to be
chosen for a competition under the condition that atleast 4 boys and atleast
4 girls must be represented. The 2 girls who won the prizes last year should be
included. In how many ways can the selection are made?
Solution:
There are 12 boys and 10 girls. From these we have to select 10 students.
Since two girls who won the prizes last year are to be included in every
selection.
95
So, we have to select 8 students from 12 boys and 8 girls, choosing atleast
4 boys and atleast 2 girls. The selection can be formed by choosing
(i) 6 boys and 2 girls
(ii) 5 boys and 3 girls
(iii) 4 boys and 4 girls
.•. Required number of ways = (12C6 x 8^2) + (12C5 x 8C3) + (12C4 x 8^4)
= (924 x 28) + (792 x 56) + (495 x 70)
= 25872 + 44352 + 34650
= 104874
Example 3.44: How many diagonals are there in a polygon?
Solution: A polygon of n sides has n vertices. By joining any two vertices
of a polygon, we obtain either a side or a diagonal of the polygon.
Number of line segments obtained by]
joining the vertices of a n sided f=Number of ways of selecting 2 out of n
polygon taken two at a time J
_ n(n  1)
— nS2 — 2
Out of these lines, n lines are the sides of the polygon.
n(n — 1)
.•. Number of diagonals of the polygon = ~  n
n(n  3)
2
Example 3.45 How many different sections of 4 books can be made from 10
different books, if (i) there is no restriction
(ii) two particular books are always selected;
(iii) two particular books are never selected?
Solution:
10!
(i) The total number of ways of selecting 4 books out of 10 = ioC4= 4 , fi , = 210
(ii) If two particular books are always selected.
This means two books are selected out of the remaining 8 books
8!
.". Required number of ways = 8C2 = tTaT = 28
(iii) If two particular books are never selected
This means four books are selected out of the remaining 8 books.
8!
.". Required number of ways = 8C4 = ~T\ 41 = 70
96
Example 3.46:
In how many ways players for a cricket team can be selected from a
group of 25 players containing 10 batsmen, 8 bowlers, 5 allrounders and 2
wicket keepers? Assume that the team requires 5 batsmen, 3 allrounder, 2
bowlers and 1 wicket keeper.
Solution:
The selection of team is divided into 4 phases:
(i) Selection of 5 batsmen out of 10. This can be done in 10C5 ways,
(ii) Selection of 3 allrounders out of 5. This can be done in 5C3 ways.
(iii)Selection of 2 bowlers out of 8. This can be done in 8C2 ways.
(iv)Selection of one wicket keeper out of 2. This can be done in 2C1 ways.
.". The team can be selected in 10C5 x 5C3 x gC2 x 2C4 ways
= 252 x 10 x 28 x 2 ways
= 141120 ways
Example 3.47: Out of 18 points in a plane, no three are in the same straight
line except five points which are collinear. How many
(i) straight lines (ii) triangles can be formed by joining them?
Solution:
(i) Number of straight lines formed joining the 18 points,
taking 2 at a time = I8C2 =153
Number of straight lines formed by joining the 5 points,
taking 2 at a time = 5C2 =10
But 5 collinear points, when joined pairwise give only one line.
.". Required number of straight lines = 153  10 + 1 = 144
(ii) Number of triangles formed by joining the 18 points,
taken 3 at a time = 18C3 = 816
Number of triangles formed by joining the 5 points,
taken 3 at a time = 5C3 = 10
But 5 collinear points cannot form a triangle when taken 3 at a time.
.". Required number of triangles = 81610 = 806
EXERCISE 3.5
1. If there are 12 persons in a party, and if each two of them shake hands
with each other, how many handshakes happen in the party?
97
2. In how many ways a committee of 5 members can be selected from
6 men and 5 women, consisting of 3 men and 2 women?
3. How many triangles can be obtained by joining 12 points, five of, which
are collinear?
4. A box contains 5 different red and 6 different white balls. In how many
ways 6 balls be selected so that there are atleast two balls of each colour?
5. In how many ways can a cricket team of eleven be chosen out of a batch
of 15 players if
(i) there is no restriction on the selection
(ii) a particular player is always chosen;
(iii) a particular player is never chosen?
6. A candidate is required to answer 7 questions out of 12 questions which
are divided into two groups, each containing 6 questions. He is not
permitted to attempt more than 5 questions from either group. In how
many ways can he choose the 7 questions.
7. There are 10 points in a plane, no three of which are in the same straight
line, excepting 4 points, which are collinear. Find the
(i) the number of straight lines obtained from the pairs of these points
(ii) number of triangles that can be formed with the vertices as these
points.
8. In how many ways can 21 identical books on Tamil and 19 identical
books on English be placed in a row on a shelf so that two books on
English may not be together?
9. From a class of 25 students, 10 are to be chosen for an excursion party.
There are 3 students who decide that either all of them will join or none
of them will join. In how many ways can they be chosen?
3.4 Mathematical Induction:
Introduction:
The name 'Mathematical induction' in the sense in which we have given
here, was first used by the English Mathematician Augustus DeMorgan
(1809  1871) in his article on 'Induction Mathematics' in 1938. However the
originator of the Principle of Induction was Italian Mathematician Francesco
Mau Rolycus (1494  1575). The Indian Mathematician Bhaskara (1153 A.D)
had also used traces of 'Mathematical Induction' in his writings.
"Induction is the process of inferring a general statement from the truth of
particular cases".
98
For example, 4 = 2 + 2, 6 = 3 + 3, 8 = 3 + 5, 10 = 7 + 3 and so on.
From these cases one may make a general statement "every even integer
except 2 can be expressed as a sum of two prime numbers. There are hundreds
of particular cases where this is known to be true. But we cannot conclude that
this statement is true unless it is proved. Such a statement inferred from
particular cases is called a conjecture. A conjecture remains a conjecture until it
is proved or disproved.
Let the conjecture be a statement involving natural numbers. Then a method
to prove a general statement after it is known to be true in some particular cases
is the principle of mathematical induction.
Mathematical induction is a principle by which one can conclude that a
statement is true for all positive integers, after proving certain related
propositions.
The Principle of Mathematical Induction:
Corresponding to each positive integer n let there be a statement or
proposition P(n).
If (i) P(l)istrue,
and (ii) P(k + 1) is true whenever P(k) is true,
then P(n) is true for all positive integers n.
We shall not prove this principle here, but we shall illustrate it by some
examples.
Working rules for using principle of mathematical induction:
Step (1) : Show that the result is true for n = 1.
Step (2) : Assume the validity of the result for n equal to some
arbitrary but fixed natural number, say k.
Step (3) : Show that the result is also true for n = k + 1 .
Step (4) : Conclude that the result holds for all natural numbers.
2
Example 3.48: Prove by mathematical induction n + n is even.
2
Solution: Let P(w) denote the statement "n + n is even"
Step (1):
Put n = 1
2 2
n + n = 1+1
= 2, which is even
.'. P(l) is true
Step (2):
Let us assume that the statement be true for n = k
99
(i.e.) assume P(k) be true.
2
(i.e.) assume "k + k is even" be true ... (1)
Step (3):
To prove P(k + 1) is true.
2
(i.e.) to prove (k + 1) + (k + 1) is even
Consider (k + l) 2 + (k + 1) = £ 2 + 2k + 1 + fc + 1
= k 2 + 2k + k + 2
= (k 2 + k) + 2(k + 1)
= an even number + 2(k + 1), from (1)
= sum of two even numbers
= an even number
.. P(k + 1) is true.
Thus if P(k) is true, then P(k + 1) is also true.
Step (4):
.". By the principle of Mathematical induction, P(w) is true for all weN.
2
i.e. n + n is even for all weN.
n(n + 1)
Example 3.49: Prove by Mathematical induction l+2 + 3 + ...+ « = j >
weN
nin + 1)
Solution: Let P(w) denote the statement : "l+2 + 3 + ... + « = ~ "
Put n = 1
P(l) is the statement : 1 = ~
1(2)
2
1 = 1
1 =
.'. P(l) is true
Now assume that the statement be true for n = k.
(i.e.) assume P(k) be true.
k(k+ 1)
(i.e.) assume I + 2 + 3 + ... + k = — ~ — • • • (1) b e true
To prove P(fc + 1) is true
(i.e.) to provel +2 + 3 + ... + k + (k+ 1)= ~ i s true '
k(k + 1)
[1+2 + 3 + ... + k] + (k+l) = 2 ' + (k+l) from(l)
100
k(k + 1) + 2(k + 1)
2
(k+l)(k + 2)
2
:. P(k + 1) is true.
Thus if P(k) is true, then P(k + 1) is true.
By the principle of Mathematical induction, P(n) is true for all neN
n(n + 1)
.. 1+2 + 3 + ... + /!= 2 forallneN
Example 3.50: Prove by Mathematical induction
,2 2 .2^ 2 w(w + 1) (2w + 1)
1 +2 +3 +...+n = t forallneN
Solution:
Let P(n) denote the statement "1 +2 +3 +...+« = 7 "
Put n = 1
2 1(1 + 1) [2(1) +1]
P(l) is the statement : 1 = 7
. K2) (3)
1  6
1 = 1
.". P(l) is true.
Now assume that the statement be true for n = k.
(i.e.) assume P(k) be true.
... ,2^,2 .2^ .2 k(k+l)(2k+l)
(i.e.) 1 +2 +3 +...+& = t ■•■(!)
To prove : P(k + 1) is true
,. . ,2 2 2 ,2 ., ,,2 (k+l)(k + 2)(2k + 3) .
(i.e.) to prove: 1 +2 +3 +...+K +(fe+l) = t is true.
[l 2 + 2 2 + 3 2 + ... +fc 2 ] + (^l) 2 = ^ +1) 6 (2/c+1) + (^ + l) 2
k(k + I) (2k +1) + 6(k + I) 2
6
_ (k + I) [k(2k + I) + 6(k + I)]
6
(yt + 1) (2/t 2 + 7/t + 6)
101
i 2 ^9 2 ^ 2 ^ a 2 .^n 2 (k+l)(k + 2)(2k + 3)
1 +2 +3 +...+K +(k+1) = 7
.. P(& + 1) is true
Thus if P(k) is true, then P(k + 1) is true.
By the principle of Mathematical induction, P(n) is true for all neN
r ^ ! 2 ^o 2 ^ ^ 2 "(w+l)(2w+l) r „ M
(i.e.) 1 + 2 + ...+« = t for all neN
Example 3.51: Prove by Mathematical induction
l2 + 23 + 3A + ... + n(n + l) = n(n+1) 3 (n + 2 \ neN.
Solution:
n(n + 1) (n + 2)
Let P(n) denote the statement "1.2+2.3 + 3.4 +. . .+ n(n + 1)= — y "
Put « = 1
P(l) is the statement : 1(1 + 1) = 3
K2) = ^P
,2(3)
z_ 3
2 = 2
.". P(l) is true.
Now assume that the statement be true for n = k.
(i.e.) assume P(k) be true
k(k + 1) (k + 2)
(i.e.) assume 1.2 + 2.3 + 3.4+...+ k(k+ 1) = — 3 L be true
To prove : P(k + 1) is true
i.e. to prove :
1.2 + 2.3 + 3.4+...,+ ^+l) + (^+l)(^ + 2) = (/c+1)(/C 3 2)(/c + 3)
Consider 1.2 + 2.3 + 3.4 + . . .,+ k(k + 1) + (k + 1) (k + 2)
= [l.2 + 2.3 + ...+fc(fc+l)] + (fc+l)(fc + 2)
fc(fc+lHfc+2) „ 1W , „ N
= — y + (k + 1) (k + 2)
fc(fc+l)(fc + 2) + 3(fc+l)(fc + 2)
3
(&+!)(* + 2) (ft + 3)
3
102
.. P(k + 1) is true
Thus if P(k) is true, P(k + 1) is true.
By the principle of Mathematical induction, P(n) is true for all neN.
, „ „ , „ „ , ^ re(n + 1) (n + 2)
1.2 + 2.3 + 3.4+ ... +n(n+ 1) =— y L
3«
Example 3.52: Prove by Mathematical induction 2  1 is divisible by 7, for
all natural numbers n.
Solution:
3«
Let P(n) denote the statement "2  1 is divisible by 7"
Put n = 1
Then P(l) is the statement : 2 3(1)  1 = 2 3  1
= 81
= 7, which is divisible by 7
.". P(l) is true
Now assume that the statement be true for n = k
3k
(i.e.) assume P(k) be true, (i.e.) "2  1 is divisible by 7" be true
Now to prove P(fc + 1) is true, (i.e.) to prove 2 A  1 is divisible by 7
Consider
2 Xk+l)_
l=2 3fe+3 l
= 2 3k . 2 3  1 = 2 3fc . 8  1
3£
= 2 .81+88 (add and subtract 8)
3k
= (2  1) 8 + 8  1
3k
= (2  1) 8 + 7 = a multiple of 7 + 7
= a multiple of 7
:. 2  1 is divisible by 7
.. P(& + 1) is true
Thus if P(k) is true, then P(k + 1) is true.
By the principle of Mathematical induction, P(n) is true for all we N
3«
:. 2  1 is divisible by 7 for all natural numbers n.
n n
Example 3.53: Prove by Mathematical induction that a  b is divisible by
(ab) for all n e N
n n
Solution: Let P(w) denote the statement "a  b is divisible by a  b".
103
Put n = 1
Then P(l) is the statement : a b = a  b is divisible by a  b
:. P(l) is true.
Now assume that the statement be true for n = k.
k k
(i.e.) assume P(k) be true, (i.e.) a  b is divisible by (a  b) be true.
a  b
=> — = c (say) where ceN
a  b J
=> a  b = c(a  b)
k k
=> a = b +c(ab) ... (1)
k + I k+ 1
Now to prove P(k + 1) is true, (i.e.) to prove : a  b is divisible
by a  b
Consider a  b = a . a b b
= [b k + c(a  b)] ab b
k k
= b a + ac(a  b)  b b
k
= b (a — b) + ac (a — b)
k
= (ab)(b + ac) is divisible by (a  b)
:. P(k + 1) is true.
By the principle of Mathematical induction, P(w) is true for all we N
n n
:. a  b is divisible by a  b for all n e N
EXERCISE 3.6
Prove the following by the principle of Mathematical Induction.
(1) (In + 1) (7m  1) is an odd number for all tie N
(2) 2 + 4 + 6 + 8 + . . . + In = n (n + 1)
(3) l + 3 + 5+...+(2«l) = « 2
(4) 1+4 + 7+ ...+(3n2)= " 2 ~ 
(5) 4 + 8+12+...+4w = 2w(«+l)
,~ ,3 .3 ,3 3 n 2 (n + l) 2
(6) 1 +2 +3 + ... +n = ^
,_. I J_ J_ J_ . J_
(7) 2 + 9 2 + 7? + ■■■ + i n ~ 2"
104
(8) In the arithmetic progression a, a + d, a + 2d, . . .
i th . . . . ,
then term is a + (n  \)d
2n
(9) 5  1 is divisible by 24 for all mg N
(10) 10 2 " ~ l + 1 is divisible by 1 1.
(1 1) n(n + 1) (n + 2) is divisible by 6 where n is a natural number.
3 2
(12) ThesumS„ = « + 3« + 5n + 3 is divisible by 3 for all n e N
(13) 7 2 " + 16«  1 is divisible by 64
(14) 2" > w for all we N
3.5 Binomial Theorem:
Introduction:
A BINOMIAL is an algebraic expression of two terms which are
connected by the operation '+' (or) ''
3
For example, x + 2y, x  y, x +4y, a + b etc.. are binomials.
Expansion of Binomials with positive Integral Index:
We have already learnt how to multiply a binomial by itself. Finding
squares and cubes of a binomial by actual multiplication is not difficult.
But the process of finding the expansion of binomials with higher powers
10 17 25
such as (x + a) , (x + a) ,(x+ a) etc becomes more difficult. Therefore we
look for a general formula which will help us in finding the expansion of
binomials with higher powers.
We know that
(x + a) =x + a = jCrj x a + jCi x a
,2 2 . 2 r 20 . 11 „ 02
(x + a) =x +2ax + a = 2^0% a + iy.\x a + 2^2* a
33 2 23 30 21 12 03
(x + a) =x + 3x a + 3xa +a = 3C0* a + j,C\x a + 3C2* a + 3C3X a
44322 34 40 31 22 13 04
(x+a) =x +4x +6x a +4xa +a =<\C<yc a +4C1X a +4C2X a +4C3X a +4C4X a
n
For n = 1, 2, 3, 4 the expansion of (x + a) has been expressed in a very
systematic manner in terms of combinatorial coefficients. The above
n
expressions suggest the conjecture that (x + a) should be expressible in the
form,
105
, ,« „ « _ n — 1 1 „ 1 «  1 _ «
(x + a) = nto i a + nL\x a + ...+nC M _ix a + nLnxa
In fact, this conjecture is proved to be true and we establish it by using
the principle of mathematical induction.
Theorem 3.6: (Binomial theorem for a Positive Integral Index)
Statement: For any natural number n
. .n „ n „ n  1 1 „ n r r ,
(x + a) = wCo x a + nC ix a + . . . + nLr x a + ...
„ ln1 «
+ wC M _ \x a + nLn x a
Proof:
We shall prove the theorem by the principle of mathematical induction.
Let P(w) denote the statement :
. .« _ « _ n1 1 ^ rcr r
(x + a) = uLqx a + mC]x a +... + nLr x a + ...
„ 1 n  1 „ «
+ «C M _ i x a + nLn x a
Step (1) :
Put n = 1
Then P(l) is the statement : (x + a) = iQ>x a + 1C1 x a
x + a = x + a
.'. P(l) is true
Step (2):
Now assume that the statement be true for n = k
(i.e.) assume P(k) be true.
, .k ,„ k , „ fe1 1 ,_, fc2 2 ,_, kr r _, Ofe
(x + a) =/c(_ox a +/cC]x a +/cL2X a +...+ kL r x a +...+ £(_£x a
be true ... (1)
Step (3):
Now to prove P(k + 1) is true
(i.e.) To prove:
.K+l _ fc+1 _ (jfc+l)ll _ (*+l)2 2
(x + a) = (jfc+l)Co* +(Jt+i)Cix a+(jt+i)C2x a +...
_ (fc+ l)r r _ fe+1
+ (fc+l)C,x a +... + (jfc+i)C(jfc+i)a
fe+ 1 A:
Consider (x + a) = (x + a) (x + a)
rI _ jfe ,„ i11 ,„ k 2 2 ._ fe(rl) (r1)
= [kCqx +/cCix a +/cC2X a + ... + kC( r _i)X a
+ kC r x a + ... + kC^a ] (x + a)
106
r .„ k+\ k \ k12 kr+2 r1
= [kL.QX + kL\x a + W_2X a + ... + kL. r \x a
+ kC r x a + ... + kCfcxa ]
r ,„ k , „ k 1 2 ,_ t2 3 ,„ kr+lr
+ [kLqx a + kL\x a +kL2X a + ...+ kL r _ix a
,_ k  r r+\ ,^ fc+ 1,
+ kL. r x a + . . . + kLkCi J
k + 1 k+ 1 fe fe12
(x + a) = fcCox + (kC{ + kCo) x .a + (kC2 + kC[) x a
+ ... + (kC r + kC r  \)x a+...+kCka •••(2)
We know that kC r + kC r  \ = (k+ 1)0
Put r= 1,2,3, ... etc.
kC\ +kCo = (fc+i)Ci
&C2 + fcCi = (fc+i)C2
fcC r + &C r i = ( t + i)C r for 1 < r < k
kCo = 1 = (fc+i)Co
fcCfe = 1 = (jfc+i)C(fc+l)
.". (2) becomes
. . fc + 1 ^ fc + 1 ^fc „ 112
(x + a) = (jt+1) Cox + (fe+ i)Cix a + (fe+i)C2X a
_ fc+ 1  r r _ yt+1
+ •■• +(k+l)C r x a + ... + (Jt+i)C(jt+i)a
.. P(£ + 1) is true
Thus if P(&) is true, P(& + 1) is true.
.". By the principle of mathematical induction P(re) is true for all neN
. .« _ « _ n — 1 1 ^ n r r
(x + a) = uLqx a + nL\x a +... + reCrx a + ...
_, 1 re  1 _, n , „ ..
+ reC M _ i x a + reCre x a tor all n e N
Some observations:
1. In the expansion
. .« _ « _ re — 1 1 ^ nr r
(x + a) = uLqx a + nL\x a +... + reCrx a + ...
_ 1 re — 1 _ « . , • ^ n r r
+ reC M _ i x a + reCre x a , the general term is nL r x a .
Since this is nothing but the (r + 1) term, it is denoted by T r + i
r + i = nL r x a .
2. The (re + 1) termisT n+ i = nC n x n ~ n a n = nC n a n , the last term.
Thus there are (n + 1) terms in the expansion of (x + a) n
107
3. The degree of x in each term decreases while that of "a" increases such
that the sum of the powers in each term is equal to n.
n
We can write (x + a) = £ nC r x a
r=0
4. hCq, «Ci, nC2, • ••, nC r , ... , nC n axe called binomial coefficients. They
are also written as Co, Ci , C2, ■ • • , C n .
5. From the relation nC r = nC n _ r , we see that the coefficients of terms
equidistant from the beginning and the end are equal.
6. The binomial coefficients of the various terms of the expansion of
n
(x + a) for n = 1, 2, 3, . . . form a pattern.
Binomials Binomial coefficients
(x + a) l
(x + a) ll
(x + a) 2 l 2 l
(x + a) 3 l 3 3 l
(x + a) 4 l 4 6 4 l
(x + a) 5 l 5 10 10 5 l
This arrangement of the binomial coefficients is known as Pascal's
triangle after the French mathematician Blaise Pascal (1623  1662). The
numbers in any row can be obtained by the following rule. The first and last
numbers are l each. The other numbers are obtained by adding the left and right
numbers in the previous row.
I, 1 + 4 = 5, 4 + 6 = 10, 6 + 4=10, 4+1=5, 1
Some Particular Expansions:
In the expansion
. .« _ n _ n 1 1 _ n r r
(x + a) =uLqx a + nL\x a +...+ nLrx a +...
„ 1 n  1 „ n .,,
+ riL n  1 x a + nLn x a ... (1)
1. If we put  a in the place of a we get
n n nlln2 2
:.(x — a) = nLox nL\x a + nL2x a ...
, 1x r _, n  r r , ,,« „ n
+ (1) nL r x a +... + (— 1) nL n a
108
We note that the signs of the terms are positive and negative
alternatively.
2. If we put 1 in the place of a in (1) we get,
Yl 2 Y Yl
(1 + x) = 1 + nC\x + nC2X +...+nC r x +...+nC n x •••(2)
3. If we put  x in the place of x in (2) we get
Yl 2 X Y Yl Yl
(1 — x) =lnC\x + nC2X ... + (1) nC r x +...+(1) nC n x
Middle Term:
Yl
The number of terms in the expansion of (x + a) depends upon the index n.
The index is either even (or) odd. Let us find the middle terms.
Case (i) : n is even
The number of terms in the expansion is (n + 1), which is odd.
Therefore, there is only one middle term and it is given by T«
2 +1
Case (ii) : n is odd
The number of terms in the expansion is (n + 1), which is even.
Therefore, there are two middle terms and they are given by T« + 1 and
2
Tn + 3
2
Particular Terms:
Sometimes a particular term satisfying certain conditions is required in
Yl Yl
the binomial expansion of (x + a) . This can be done by expanding (x + a) and
then locating the required term. Generally this becomes a tedious task, when the
index n is large. In such cases, we begin by evaluating the general term
T r+ i and then finding the values of r by assuming T r+ i to be the required term.
To get the term independent of x, we put the power of x equal to zero and
get the value of r for which the term is independent of x. Putting this value of
r in T r+ i, we get the term independent of x.
5 ( 3^ 4
Example 3.54:Find the expansion of : (i) (2x + 3y) (ii) ( 2x — — \
Solution:
(i) (2x + 3y) 5 = 5 C (2x) 5 (3yf + 5C1 (2x) 4 (3y) 1 + 5 C 2 (2x) 3 (3y) 2
+ 5 C 3 (2x) 2 (3y) 3 + 5C4 (2x) 1 (3y) 4 + 5C5 (2x)° (3y) 5
= l(32)x 5 (1) + 5(16x 4 ) (3;y) + 10(8x 3 ) (9y 2 )
109
+ 10(4jc 2 ) (21y 3 ) + 5(2x) (8l/) + (1) (1) (243/)
= 32jc 5 + 240x 4 y + 720x 3 y 2 + 1080xV + 810xy 4 + 243y 5
(Hi \2x 2 ^] = 4 C (2x 2 ) 4 (£) + 4 Ci(2x 2 ) 3 ()
2 N 2 ( 3^ 2 „ _ 2 N 1 T 3V ^ ,~1S>( 3^ 4
+ 4 C 2 (2x) I J + 4 C 3 (2x) {) + 4 C 4 (2x) v v
= (1) 16x 8 (l) + 4(8x 6 ) ( 1) + 6(4/) (J) + 4(2* 2 ) [ j
+ (D(1)^
= 16jc 8  96x 5 + 216jc 2  — + ~j
x
Example 3.55: Using binomial theorem, find the 7 power of 1 1 .
Solution:
ll 7 = (1 + 10) 7
= 7C0 (l) 7 (10)°+ 7 Ci (l) 6 (10) 1 + 7 C2(1) 5 (10) 2 +7C 3 (1) 4 (10) 3 +7C 4 (1) 3 (10) 4
+ 7C5 (l) 2 (10) 5 + 7 C 6 (l) 1 (10) 6 + 7C7 (D° (10) 7
7x6 2 7x6x5 37x6x5 4 7x6 5 6 7
= 1+70+— rlO + . .  10 +  . . 10 4 + r 10 +7(10)° + 10 7
1x2 1x2x3 1x2x3 1x2
= 1 + 70 + 2100 + 35000 + 350000 + 2100000 + 7000000 + 10000000
= 19487171
r \ X1
Example 3.56: Find the coefficient of x in the expansion of \x + ~~y
I x
Solution:
n 17
In the expansion of [ x + 3 , the general term is
x )
T r+ i = nC r x r j I
r 17 4r
Let T r + 1 be the term containing x
then, 174r=5 =^> r = 3
110
•'• Tr+ 1 = T3+ i
174(3) 5
= 17C3 x K ' = 680x
.". coefficient of x = 680
( 2V
Example 3.57: Find the constant term in the expansion of K/x  ~~2
\ x )
Solution:
10
In the expansion of [ yfx  ~j
T r +i =
10Cr (V*)
 r
K x )
=
10 r
10C r x 2
2r  10Cr ( 2) x
10 5r
2r
=
K)C r (2) x
2
Let T r + 1 be the constant term
Then,
10 5r
2
=> r=2
105(2)
l C 2 (2) x 2
.". The constant term

=
10x9 .
. x 4 xx
1x2

18
f7
Example 3.58: If we N, in the expansion of (1 + x) prove the following :
n
(i) Sum of the binomial coefficients = 2
(ii) Sum of the coefficients of odd terms = Sum of the coefficients of even
r,n 1
terms = 2
Solution: The coefficients «Co, nC[, nQ% ... , nC n in the expansion of
ft
(1 + x) are called the binomial coefficients, we write them as Co, Ci, C2, • • • C n ,
n 2
(1+x) = C0 + C1X + C2X +...+■
It is an identity in x and so it is true for all values of x.
II 2 T Yl
(1+x) = C0 + C1X + C2X +...+C r x +...+C n x
111
Putting x = 1 we get
2 n = C + Ci+C 2 +...+C w ...(1)
put x =  1
o = CoCi + c 2 c 3 + ...(i) n c n
=^Q) + C2 + C4+ ...= C1+C3 + C5+...
It is enough to prove that
C0 + C2 + C4+... = C1+C3 + C5+ ... =2 n ~ l
Let C0 + C2 + C4+... = C1+C3 + C5+... =k ... (2)
n
From(l), C0 + C1+C2+ ... +C„ = 2
2k = 2 n From (2)
k =2 n ~ l
From (2), C0 + C2 + C4+... = Ci + C3 + C5 + ... = 2" ~ !
EXERCISE 3.7
(1) Expand the following by using binomial theorem
(i) (3a + 5b) 5 (ii) (a  2b) 5 (iii) (2x  3x 2 ) 5
(iv) [x + J (v) (x +2y ) (vi) (x Jy + yyjx)
(2) Evaluate the following:
(i) (V2 + l) 5 + (V2  l) 5 (ii) (V5 + l) 5  ( V3  l) 5
(iii) ( 1 + V5) 5 + (1  ^/5) 5 (iv) ( 2sfa + 3) 6 + ( 2y[a  3) 6
(v)(2 + V3) 7 (2a/3) 7
3 3
(3) Using Binomial theorem find the value of (101) and (99) .
3
(4) Using Binomial theorem find the value of (0.998) .
(5) Find the middle term in the expansion of
2x 2 r fb xv 6
(i) \3x^r (ii) 7 +
3 ; w u
16 r i\17
x
m(,^i\ (iv)(.v2y) B (v)f* + 4
*)
112
(6) Show that the middle term of
2 „ 1.3.5.7 ...(2« 1)2"/
(i)(l+x) is ~ {
....( if" . 1.3.5. ...(2W1)
(ii)(x + ^J is ~ {
f if" . (l) W .1.3.5.7....(2nl)
< m > {* ~ x) 1S n\ 2
11
(7) Find the coefficient of x in the expansion of I x — J
(8) Find the term independent of x (constant term) in the expansion of
( 2 if 2 (4x 2 3 f ( b X1
(H 2 * + x) ^{—Tx) W{ 9X ^
(9) In the expansion of (1 + x) , the coefficient of r and (r + l) 1 terms are
in the ratio 1 : 6, find the value of r.
If the coefficients
are in A. P., find n.
(10) If the coefficients of 5 ,6 and 7 terms in the expansion of (1 + x)
113
4. SEQUENCE AND SERIES
4.1 Introduction
We hear statements such as "a sequence of events", "a series of tests before
the board examination", "a cricket test match series". In all these statements the
words "sequence" and "series" are used in the same sense. They are used to
suggest a succession of things or events arranged in some order. In mathematics
these words have special technical meanings. The word 'sequence' is used as in
the common use of the term to convey the idea of a set of things in order, but
the word "series" is used in a different sense.
Let us consider the following example.
A rabbit and a frog are jumping on the same direction. When they started
they were one metre apart. The rabbit is jumping on the frog in order to catch it.
At the same time the frog is jumping forward half of the earlier distance to
avoid the catch. The jumping process is going on. Can the rabbit catch the frog?
1
IS
«l I a, 1
R i * ' F,  R, j2_F,yR,
• • • • ., »
1 R* 22 .F 2 ± r^f 4
Fig. 4. 1
Let a\, a% 03, a\ ... be the distances between the rabbit and the frog at the
first, second, third, fourth instants etc,. The distance between the rabbit and the
frog at the first instant is 1 metre.
1 1 J_ 1 J_
.. a\ = 1 ; (22 = 2 ; a 3 = 4 = 2 ; a 4 = 3 = 3
Here a\, a% ai, ... form a sequence. There is a pattern behind the
arrangement of a\, a2, 03 ... Now a n has the meaning,
(i.e.) a n is the distance between the rabbit and the frog at the n instant
When a n becomes the rabbit will catch the frog.
i.e. the distance between the frog and the rabbit is zero when n —> go
Further
a n 
1
"2"
1
As n >
■oc,
a,,
>
114
At this stage the rabbit will catch the frog.
This example suggests that for each natural number there is a unique real
number,
i.e. 1 2 3 ... n
V* 4* V* V*
a i ci2 03 ... a n
1111 1
= 1
2 2 1 4 2 2 2""
Consider the following list of numbers
(a) 8, 15, 22, 29, (b) 6, 18, 54, 162,
In the list (a) the first number is 8, the 2 n number is 15, the 3 r number is
22, and so on. Each number in the list is obtained by adding 7 to the previous
number.
In the list (b) the first number is 6, the 2 n number is 18, the 3 r number is
54 etc. Each number in the list is obtained by multiplying the previous number
by 3.
In these examples we observe the following:
(i) A rule by which the elements are written (pattern).
(ii) An ordering among the elements (order).
Thus a sequence means an arrangement of numbers in a definite order
according to some rule.
4.2 Sequence
A sequence is a function from the set of natural numbers to the set of real
numbers.
If the sequence is denoted by the letter a, then the image of n e N under
the sequence a is a(n) = a n .
Since the domain for every sequence is the set of natural numbers, the
images of 1, 2, 3, ... n .. . under the sequence a are denoted by a\, a2, a^ ... a n ,
. . . respectively. Here a\, a2, aj, ...a n , ... form the sequence.
"A sequence is represented by its range".
Recursive formula
A sequence may be described by specifying its first few terms and a
formula to determine the other terms of the sequence in terms of its preceding
terms. Such a formula is called as recursive formula.
115
For example, 1, 4, 5, 9, 14, . . ., is a sequence because each term (except the
first two) is obtained by taking the sum of preceding two terms. The
corresponding recursive formula is a n + 2 = a n + a n + l . n  1 here «i=l, «2= 4
Terms of a sequence:
The various numbers occurring in a sequence are called its terms. We
denote the terms of a sequence by a\, a2, aj,, ... , a n , ... , the subscript denote
the position of the term. The n term is called the general term of the sequence.
For example, in the sequence 1, 3, 5, 7, ... 2«  1, ...
the 1 st term is 1, 2 n term is 3, and n l term is 2n  1
Consider the following electrical circuit in which the resistors are indicated
with sawtoothed lines.
"■Wv
8
3
— ^•V'N
"
Fig. 4. 2
If all the resistors in the circuit are 1 ohm with a current of 1 ampere then
the voltage across the resistors are 1, 1, 2, 3, 5, 8, 13, 21, ...
In this sequence there is no fixed pattern. But we can generate the terms of
the sequence recursively using a relation. Every number after the second is
obtained by the sum of the previous two terms,
i.e. Vi = 1
V 2 = 1
v 4 = v 3 + v 2
v 5 = v 4 + v 3
V = V
'I? — v n
+ V„
116
Thus the above sequence is given by the rule:
Vi = l
V2 = 1
Vn = Vn  1 + Vn  2 ; n > 3
This sequence is called Fibonacci sequence. The numbers occurring in this
sequence are called Fibonacci numbers named after the Italian Mathematician
Leonardo Fibonacci.
Example 4.1:
Find the 7 C term of the sequence whose n term is ( 1)" +
Solution:
,n + l (n + 1
n + 1
n
substituting n = 1, we get
n7 + 1
Given a n = ( 1)
n = 1 , we
ai = ( l) 7 K j]  7
4.3 Series
For a finite sequence 1, 3, 5, 7, 9 the familiar operation of addition gives
the symbol 1+3 + 5 + 7 + 9 which has the value 25.
If we consider the infinite sequence 1, 3, 5, 7, ... then the symbol
1 + 3 + 5 + 7+.. . has no definite value, because when we add more and more
terms the value steadily increases. 1 + 3 + 5 + 7 + 9+.. .is called an infinite
series. Thus a series is obtained by adding the terms of a sequence.
If a\, ai, aj,, ... a n ... is an infinite sequence then a\+ a2+ ... + a n + ... is
00
called an infinite series. It is also denoted by X a k
k=\
If S„ = a\ +^2+ • • • + a n then S„ is called the n l partial sum of the series
oo
k=l
00 i
Example 4.2 Find the n l partial sum of the series X ~
n=l 2
Solution:
J_ J_ J_
S„ 2l + 22 +...+ 2 „
117
11 1 1
and S„ + i =^j +^ +...+— +^TT
$n+ 1 = S„ + „+ i
Also we can write S„ + i as
J_ J_ J_ 1
S« + l  i + 2 2 + • • • + 2 « + 2 n+[
1 1
1
...(1)
S n+1 =2[l+S n ] ...(2)
From (1) and (2) S„ + —[ = 2^ + S «l
Note: This can be obtained by using the idea of geometric series also. We know
q(lr ,T )
that the sum to n terms of a geometric series is S„ = "
(1r)
Here a = 9 , n = n, r = ~z (< 1)
^n ~~
^2
n«
■§
1 2"
EXERCISE 4.1
(1) Write the first 5 terms of each of the following sequences:
r\ t n«U«+l ^ "(" + 5 )
(1) a n = ( 1) 5 (11) a n = 4
(iii) a n =  \\n + 10
1 / 1 \« 2
1  ( 1) M
(v) «„ = 3 (vi) a n = —
118
(2) Find the indicated terms of the following sequences whose n term is
1 (n%\
(i)a n = 2+ ; a 5 , a 7 (n) a„ = cos I y I ; a 4 , a 5
(in) an = „ ; «7 . «10 (iv)a„ = (l) 2 , a 5 , a 8
(3) Find the first 6 terms of the sequence whose general term is
2
n — 1 if « is odd
a„ =
« 2 +l ., .
it « is even
2
(4) Write the first five terms of the sequence given by
(i) a\ = a2 = 2, a n = a n _ i  1, n > 2
(ii) a\ = \, ci2 = 2, a n = a n _ i + a n _ 2, n > 2
(iii) a\ = 1 , a n = «a„ _ 1 , « > 2
(iv) ai = «2 = 1> a n = 2a«  1 + 3a„ _ 2, « > 2
00 1
(5) Find the n partial sum of the series X ~
n=\ 3
00
(6) Find the sum of first n terms of the series X 5"
n=\
CO 1
(7) Find the sum of 101 ! terms to 200 1 term of the series 2Z ~
n=\ 2
4.4 Some special types of sequences and their series
(1) Arithmetic progression:
An arithmetic progression (abbreviated as A.P) is a sequence of numbers in
which each term, except the first, is obtained by adding a fixed number to the
immediately preceding term. This fixed number is called the common
difference, which is generally denoted by d.
For example, 1, 3, 5, 7, ... is an A.P with common difference 2.
(2) Arithmetic series:
The series whose terms are in A.P is called an arithmetic series.
For example, 1+3 + 5 + 7+.. . is an arithmetic series.
119
(3) Geometric progression
A geometric progression (abbreviated as G.P.) is a sequence of numbers in
which the first term is nonzero and each term, except the first is obtained by
multiplying the term immediately preceeding it by a fixed nonzero number.
This fixed number is called the common ratio and it is denoted by the letter V .
The general form of a G.P. is a, ar, ar , ... , with a ^ and r * 0, the
first term is 'a'
(4) Geometric series:
2 n — \
The series a + ar + ar + ... + ar + ... is called a geometric series
because the terms of the series are in G.P. Note that the geometric series is finite
or infinite according as the corresponding G.P. consists of finite (or) infinite
number of terms.
(5) Harmonic progression:
A sequence of nonzero numbers is said to be in harmonic progression
(abbreviated as H.P.) if their reciprocals are in A. P.
The general form of H.P is , —^, —^ , ... , where a *0.
n term of H.P. is T„ = : —.
n a + (n l)d
For example the sequences 1, ■? , g , tt , ... is a H.P., since their reciprocals
1,5,9, 13, ... areinA.P.
Note: There is no general formula for the sum to n terms of a H.P. as we have
for A.P. and G.P.
Example 4.3 If the 5 th and 12 th terms of a H.P. are 12 and 5 respectively, find
the 15 th term.
Solution:
T =
1 n —
" a + (n  \)d
1 1
Given T 5 =12^ a + (5 _ l)d =12 => ^^ = 12
a + Ad = J2 •■•(!)
and Ti 2 = 5 =>   — _ j )d = 5 => —^ =5
=> a+lld = j ...(2)
120
(2)  (1)
(1) =>
ld = m
a + 4 {eo =12
j_
12
a +
60
=> rf = 60
12
60
a =
60
••• Tig =
1
1
a + (15 l)rf J +14x L
60 + 14 x 60
J_
"15
60
Ti 5 = 4
60
15
4.5 Means of Progressions
4.5.1 Arithmetic mean
A is called the arithmetic mean of the numbers a and b if and only if
a, A, b are in A. P. If A is the A.M between a and b then a, A, b are in A.P
=> Aa = b  A
=> 2A = a + b
A = 
a + b
A\, A2, •■• , A„ are called n arithmetic means between two given numbers
a and & if and only if a, A\, A2, . . . A„, & are in A.P.
Example 4.4 : Find the n arithmetic means between a and & and find their sum.
Solution:
Let A\, A2, • • • , A„ be the n A.Ms between a and b. Then by the definition
of A.Ms a, Ai, A2, ... , A n , frarein A.P
Let the common difference be cf.
.'. Ai = a + cf, A2 = a + 2c/, A3 = a + 3d, ... , A n = a + nd and b = a + (n + l)d
=> (n + Y)d = b  a
■ a
:d= n+\
.••A 1 =a + n+l
b — a
J2jb a)
A 2= fl+ n+ l
»(£>  a)
/ill — (A \ , 1
" H + 1
121
Sum of n A.Ms between a and b is
b  a
Ai+A 2 + ... +A n =
a +
n+l
a + '
2(b  a)
n+l
a +'
n(b  a)
n+l
(ba)
= na + ~~r [I +2+ ... + ri]
n + I L '
(b  a) n(n + 1) n(b  a)
= na + 77TTT\ ■ t = na + 7.
a + b
(n+l) ■ 2
2na + nb  na na + nb
2 2 =n K. 2
Example 4.5: Prove that the sum of n arithmetic means between two numbers is
n times the single A.M between them
Solution:
Let Aj, A2, ... , A„ be the n A.Ms between a and b.
From the example (4.4)
a + b
Ai + A2 + A3 + . . . + A„ = n I 2 J = n x (A.M between a and b)
= n (single A.M between a and b)
Example 4.6: Insert four A.Ms between  1 and 14.
Solution:
Let A1, A2, A3, A4 be the four A.Ms between  1 and 14.
By the definition  1, Ai, A2, A3, A4, 14 are in A. P. Let d be the common
difference.
.. A\=  1 + d, ; A 2 =  1 + 2d ; A3 =  1 + 3d, ; A 4 =  1 + Ad ; 14 = l+5d
:.d = 3
.. Ai=  1 + 3 = 2 ; A 2 =  1+2 x 3 = 5 ; A3 = 1+3x3 = 8 ; A 4 =  1 + 12 = 1 1
.. The four A.Ms are 2, 5, 8 and 1 1.
4.5.2 Geometric Mean
G is called the geometric mean of the numbers a and b if and only if
a, G, b are in GP.
G b
=> «=G =r
=> G = ab
G = ±yfab
122
Note:
(1) If a and b are positive then G = + ^[ab
(2) If a and b are negative then G =  ^Jab
(3) If a and b are opposite sign then their G.M is not real and it is
discarded since we are dealing with real sequences.
i.e. If a and b are opposite in signs, then G.M between them does not exist.
Example 4. 7: Find n geometric means between two given numbers a and b and
find their product.
Solution:
Let Gi, G2, • • • , G„ be n geometric means between a and b.
By definition a, G\, G2, ■ • • , G n , b are in G.P. Let r be the common ratio.
Then G\ = ar, G2 = ar , . . . , G n = ar n and b = ar
1
n + 1
Gi = a
V\n+1
1
The product is
V\n+1
1
G2 = a
V\n+\
G w = a
V\n + \
G\ . G2 . G3 . G n = a<
V\n+1 IU\ n + 1
V\n+\
= a
1 +2+ ... +71
n+ 1
71(71 + 1)'
V\ 2(72+1)
a)
n
= (ab) 2
Example 4.8: Find 5 geometric means between 576 and 9.
Solution:
Let Gi, G2, G3, G4, G 5 be 5 G.Ms between a = 576 and b = 9
Let the common ratio be r
Gi = 576r, G 2 = 576r 2 , G 3 = 576r 3 , G 4 = 576r 4 , G 5 = 576r 5 , 9 = 576r 6
123
r 6 =
r =
9
576
J_
2
'{576J 6 {64j 6
Gi =576r = 576x2 =288
G 3 = 576r 3 = 576 x  = 72
G 5 = 576r 5 = 576 x 32 = 18
G 2 = 576r 2 = 576 x \ = 144
G 4 = 576r 4 = 576 x jt = 36
4
J_
16
Hence 288, 144, 72, 36, 18 are the required G.Ms between 576 and 9.
Example 4.9: If b is the A.M of a and c (a ^ c) and (b  a) is the G.M of
a and c  a, show that a : b : c = 1 : 3 : 5
Solution:
Given b is the A.M of a and c
.'. a, b, c are in A. P. Let the common difference be d
:. b = a + d
c = a + 2d
Given (b  a) is the G.M of a and (c  a)
:. (b  a) = a(c  a)
2
d = a(2d)
=> d=2a [:d*0]
= a + d c = a + 2d
= a + 2a c = a + 2(2a)
...(1)
...(2)
From (1) and (2)
= 3a
c = 5a
:. a : b : c = a : 3a : 5a
=1:3:5
4.5.3 Harmonic mean
H is called the harmonic mean between a and b if a, H, b are in H.P
If a, H, b are in H.P then — , 77 , t are in A.P
1
H
1 1
a + b
1 _I I
H "a + b
124
H =
lab
a + b
This H is single H.M between a and b
Definition:
Hi, H2, ... H„ are called n harmonic means between a and b if a, Hi, H2,
. . . H„, b are in H.P.
Relation between A.M., G.M. and H.M.
Example 4.10: If a, b are two different positive numbers then prove that
(i) A.M., G.M., H.M. are in G.P. (ii) A.M > G.M > H.M
Proof:
a +
lab
A.M. = ^— ; G.M. = ^ab ; H.M. = ^
(1)
G.M _ Jab _ Isjab
A.M ~~ a + b a + b
1
lab
H.M a +
G.M
Isjab
\[ab ~ a + b
(1)
...(2)
From (1) and (2)
G.M _ H.M
A.M ~ G.M
.. A.M, G.M, H.M are in G.P
,.. N . „ „ a + b i—r a + b1Jab
(ii)A.MG.M=— j— ^Jab = 2 —
(cfafb) 2
= 2 >0 :a>0;b>0;a*b
A.M > G.M ...(1)
G.MH.M = V^ JTb
_ ^[ab (a + b) lab _ ^ab [a + b l\[ab]
~ a + b
2
>
a + ,
^[ab (^[a  ^jb)
a + b
:. G.M >H.M
From (1) and (2) A.M. > G.M > H.M
(2)
125
EXERCISE 4.2
(1) (i) Find five arithmetic means between 1 and 19
(ii) Find six arithmetic means between 3 and 17
(2) Find the single A.M between
(i) 7 and 13 (ii) 5 and  3 (iii) (p + q) and (pq)
(3) If b is the G.M of a and c and x is the A.M of a and b and y is the A.M
a c
of b and c, prove that — +  = 2
x y
(4) The first and second terms of a H.P are ~ and t respectively, find the
9 th term.
b + a b + c
(5) If a, b, c are in H.P., prove that + •; = 2
r b  a b  c
(6) The difference between two positive numbers is 18, and 4 times their
G.M is equal to 5 times their H.M. Find the numbers.
(7) If the A.M between two numbers is 1, prove that their H.M is the square
of their G.M.
2
(8) If a, b, c are in A. P. and a, mb, c are in G.P then prove that a, m b,c are
in H.P
(9) If the p and q terms of a H.P are q and p respectively, show that
(pq) 1 term is 1 .
(10) Three numbers form a H.P. The sum of the numbers is 11 and the sum
of the reciprocals is one. Find the numbers.
4.6 Some special types of series
4.6.1 Binomial series
Binomial Theorem for a Rational Index:
In the previous chapter we have already seen the Binomial expansion for a
positive integral index n. (power is a positive integer)
i , \1 n , /i n— 11, /in — 22, ,/>» — r r , ,, n
(x + a) =x + nL\ x a + nL2X a +...+ nL r x a +... + nL n a
A particular form is
„ n(n  1) 2 n ( n !)(" 2 ) 3 „
(1 +xf = 1 +nx+ 2 , x z + — jf " jc j + ... +x n
126
When n is a positive integer the number of terms in the expansion is («+l)
and so the series is a finite series. But when it is not a positive integer, the series
does not terminate and it is an infinite series.
Theorem (without proof)
For any rational number n other than positive integer
n n(n  1) 2 , U(n l)(n 2) 3
(1 + x) = 1 + nx + r^ x + i j n x +
provided I x I < I .
Here we require the condition that I x I should be less than 1 .
To see this, put x = 1 and n =  1 in the above formula for (1 + x) n
The left side of the formula = (1 + 1)~ = x ,
while the right side = 1 + ( 1) (1) + ( ~ ^  l 2 + ...
= 11 + 11 + ...
Thus the two sides are not equal. This is because, x = 1 doesn't satisfy I x I < 1.
This extra condition I x I < 1 is unnecessary, if n is a positive integer.
Differences between the Binomial theorem for a positive integral index and
for a rational index:
1. If n e N, then (1 + x) n is defined for all values of x and if n is a
rational number other than the natural number, then (1 + x) n is defined
only when I x I < I.
2. If n £ N, then the expansion of (1 + x) n contains only n + 1 terms. If
n is a rational number other than natural number, then the expansion
of (1 + x) n contains infinitely many terms.
Some particular expansions
We know that , when n is a rational index,
n , \» 1 , , w( " ~ U 2 , n(n  1) (n  2) 3
(1+x) = 1 + nx + Tj x + Tj x + ... (1)
Replacing x by  x, we get
nxrl»x + " ( "" 1) x 2  »(^ D(» 2) 3 (2)
y 1 A^ — 1 AtA T ^i A o  AT... v^v
Replacing w by  n in (1) we get
(1 +x)~ n = 1 —nx + — 2\ — x 1 — oj 1 L x 5 +... (3)
n(n + 1) 2 »(» + 1) (n + 2) 3
127
Replacing x by  x in (3), we get
,, xn , n ( n + 1) 2 »(" + 1) pi + 2) 3
(1jc) "= 1 +/uc +  2i — x + of x +... (4)
Note:
(1) If the exponent is negative then the value of the factors in the
numerators are increasing uniformly by 1
(2) If the exponent is positive then the value of the factors in the
numerators are decreasing uniformly by 1
(3) If the signs of x and n are same then all the terms in the expansion are
positive.
(4) If the signs of x and n are different, then the terms alternate in sign
Special cases
1. (1 + x) _1 = 1 x + x 2 x 3 + ...
2. (1 x)~ l = 1 +x + x 2 + x 3 + ...
3. (l+x)~ 2 = l2x + 3x 2 4x 3 + ...
4. (lx)~ 2 = l+2x + 3x 2 + 4x 3 +...
General term:
For a rational number n and I x I < 1, we have
„ n(n  1) 2 , n(n  1) (w  2) 3
(1 + X) = 1 + MX + T^ X + 1 j } x + • • •
In this expansion
First term Tj = Tq + 1 = 1
Second term T2 = Tj + \ = nx = j x
n(n  1) 7
Third term T3 = T2 + 1 = — p^ — x
_ , w(w  1) (n  2) 3
Fourth term T4 = T3 + \ = TYt, x etc 
, , n th t ^ n(n 1) (n2) ... (n(r 1)) r
(r+1) term:T r+1 = ryt x
The general term is
n(n  1) (n  2) . . ,r factors r »(»  1) (w  2)...(n r+1) r
TY+ 1 = r i x = r  x
Example 4.11: Write the first four terms in the expansions of
(i)(l+4x)" 5 wherelxl<T (ii)(lx 2 )" 4 wherelxl<l
128
Solution: (i)
I 4x I = 41 x I < 4 I 4 j = 1 .. I 4x I < 1
.". (1 + Ax) can be expanded by Binomial theorem.
(5) (5 + 1) ,. , 2 (5) (5 + D (5 + 2)
(1 + Ax)' J = 1 ( 5) {Ax) + y '\ 2 ' (Ax)
1.2.3
(4x) 3 +...
= 1  20x + 15(16x 2 )  35(64x 3 ) + . . .
= 1  20x + 2A0x 2  2240x 3 + . . .
94 2
(ii)(lx) can be expanded by Binomial theorem since I x I < 1
, A r .w ?, (4) (4+1) , 2 2 ^ (4) (4+1) (4 + 2) 2 3
= 1 + (4) (x ) + j^2 (* ) + L23 ( x ) + • • •
= 1 + 4x 2 + 10x 4 + 20x 6 + ...
Example 4.12:Find the expansion of t where Ixl < 2 upto the fourth term.
(2 + x)
Solution:
1
(2 + x)
? =(2+x)" 4 = 2" 4 1+*
\x\<2
< 1
J_
16
J_
16
l(4)
x\ , (4) (4 + 1) fx\ 2 (A) (4 + 1) (4 + 2) fx] 3
1.2
1.2.3
V2
1 2x + "
(4) (5) fx 2 ^ (4) (5) (6) .
2 U
1.2.3 8
+ ...
16 8 + 32 x 32 x +
Example 4.13:Show that (l+x) n =
2 n
\4
lx\ fn+V\(lx
i+x) + n { 2! )(l + *
?♦...]
_ . . T 1 x
Solution: Let y = ,
R.H.S = 2"
" «(«+ 1) 2 1
l/ry + 2 , y + ...
= 2 n [l+y]"
= 2"
r ix"
L 1 + i+*.
 n
= 2 n
1 +x+ 1 x
1 +x
 n
= 2"
~ 2 "
1 +x_
 n
= 2 n
"1 +x
2
n
= (l+x) n =
L.H.S.
129
Approximation by using Binomial series
Example 4.14: Find the value of ^126 correct to two decimal places.
Solution:
1 1
3,
ifl26 = (126) 3 = (125 + l) 3
1 1
J_Y
125 1 +
125/
3 = (125) 3  1 i
1
_L I 3
125 J
= 5
= 5
1 1
3
1
1 + 3 • 125 +
1
<1
+ 3 (0.008)
• 125
by neglecting other terms
= 5[1+ 0.002666]
= 5.01 (correct to 2 decimal places)
Example 4.15: If x is large and positive show thaw.* + 6  \]x + 3 =~~2 (app.)
x
Solution: Since x is large, ~ is small and hence
< 1
■\jx 3 + 6 \jx 3 + 3 = (x 3 + 6) 3  (x 3 + 3) 3 =x\l+— ] 3 x\l+— ] 3
x 3 ) "TV
1 l 6
1 + 3.3 +...
•' X
x + ^ +
f...
 X
^ + 5
3
■7 3 + 

;
1
+ ...
2
~x 2 "
J
X
2
(approximately)
1
Example 4.16: In the expansion (1  2x) 2 ) fj nc j the coefficient of jc .
Solution: We know that
(1x) =l+nx+ — 21 — x +"
»(n + 1) 2 »(» + 1) (n + 2) 3 n(n+l) ... (n + rl)
3!
X+...+ "
r!
x +.
General term T r + 1 =
w(w + 1) ... (« + r 1)
Take n = 2 anc ^ re Pl ace x by 2x.
130
1 f3\f5\ (2r\
t>.. iWW ? 2 W  1 "".^" ^
r r
r! 2' "
1.3.5 ...(2rl)
. . coerncient or jc = ~j
put r = 8
 8 1.3.5.7.9.11.13.15
.•. coerncient or x = k\
4.6.2. Exponential series
Exponential theorem (without proof)
For all real values of x,
11 1 Y x x 2 x 3
l+jy+27+...+^+...j = l+77 + 2! + 3! +
1 1 1
But e= 1 + 77 +T\ +T\ +•••
2 3
.". For all real values of x, e = 1+77+27+07 +
Thus we have the following results:
2 3
> A> •> •>
e = 1 — YT + 2T — 3T + 
x , x 2 4
e + e _ x_ x_
2 ~ + 2! + 4! + •••
X —X
e  e
2
3 5
X X
= x+ 3! + 5! + •••
e + e
2
, 1 1
= 1+2! + 4! + "
2
1 1 1
_ 1! + 3! + 5! + '
4.6.3 Logarithmic Series:
2 3 4
If  1 < x < 1 then log(l + x)=x^j~ + T"  "7" +
This series is called the logarithmic series.
131
The other forms of logarithmic series are as follows:
2 3
log(l x) = x~2 ~^  ...
2 3
x x
 log(l x)=x + y+y+...
/ 3 5
X X
log(l + X)  l0g(l  X) = 2 I X + ^ + "F +
1 1 +X X X
2 lo §r^ =x+ y + y + 
EXERCISE 4.3
(1) Write the first four terms in the expansions of the following:
(i) 7 where I x I > 2 (ii) where I x I < 2
(2 + x) 4 3r— —
V63x
(2) Evaluate the following:
o
(i) ^1003 correct to 2 places of decimals
(ii) correct to 2 places of decimals
i 2
1 — X X
(3) If x is so small show that A / , = 1  x + y (app.)
(4) If x is so large prove that ^/x + 25  Ajx +9 = ~ nearly.
11
(5) Find the 5 term in the expansion of (1  2x ) 2
(6) Find the (r + l) c term in the expansion of (1  x)
(7) Showthatx"=l+/l£) + " ( " 2 1) (l£f +...
132
5. ANALYTICAL GEOMETRY
Introduction
'Geometry' is the study of points, lines, curves, surfaces etc and their
properties. Geometry is based upon axioms and it was laid by the famous Greek
Mathematician Euclid about 300 B.C. In the 17 c century A.D., the methods of
Algebra were applied in the study of Geometry and thereby 'Analytical
Geometry' emerged out. The renowned French philosopher and Mathematician
Rene Descartes (1596  1650) showed how the methods of Algebra could be
applied to the study of Geometry. He thus became the founder of Analytical
Geometry (also called as Cartesian Geometry, from the latinized form of his
name Cartesius). To bring a relationship between Algebra and Geometry,
Descartes introduces basic algebraic entity 'number' to the basic geometric
concept of 'point'. This relationship is called 'system of coordinates'. Descartes
relates the position of a point with its distance from fixed lines and its direction.
This chapter is a continuation of the study of the concepts of Analytical
Geometry to which the students had been introduced in earlier classes.
5.1 Locus
The path traced by a point when it moves
according to specified geometrical conditions is
called the locus of the point. For example, the
locus of a point P(xj, y^) whose distance from a
fixed point C (h, k) is constant 'a', is a circle
(fig. 5.1). The fixed point 'C is called the centre
and the fixed distance 'a' is called the radius of the circle.
Example 5.1: A point in the plane moves so that its distance from (0, 1) is twice
its distance from the xaxis. Find its locus.
Solution:
Let A(0, 1) be the given point. Let
P(xj, jj) be any point on the locus. Let B be
the foot of the perpendicular from P(x 1 ,y 1 ) to
the xaxis. Thus PB =y^.
Given that PA = 2PB
P(*i,yi)
Fig. 5. 1
YT
A(0, 1)
i.e.
.. PA 2 = 4PB 2
(x l 0) 2 + (y 1 l) 2 = 4y l 2
O
P(*i,yi)
EL
B
Fig. 5. 2
133
2 2 2
i.e. Xj +)>j 2y l + l=4y l
i.e. JCj 2  3jj 2  2y l + 1 =
2 2
.". The locus of (Xp jj) is x  3j 2j + 1 =
Example 5.2: Find the locus of the point which is equidistant from ( 1, 1) and
(4,  2).
Solution:
Let A(— 1,1) and B(4,  2) be the given points.
Let P(x 1 ,y 1 ) be any point on the locus. Given that PA = PB
.. PA 2 = PB 2
i.e. (xj + 1) 2 + (y x  l) 2 = (x t  4) 2 + (^ + 2) 2
i.e. Xj + 2x 1 + l+;y 1 2y 1 + l=x 1 8x^ + 16 + )^ + 4^+4
i.e. lObCj  6y x  18 = i.e. 5x 1 3 y x  9 =
.". The locus of the point (xj, yj) is 5x  3y  9 =
Example 5.3: If A and B are the two points ( 2, 3) and (4,  5), find the
2 2
equation of the locus of a point such that PA  PB = 20.
Solution:
A(— 2, 3) and B(4,  5) are the two given points. Let P(xj, y^) be any point
on the locus. Given that PA 2  PB 2 = 20.
(xj + 2) 2 + (y 1  3) 2  [(xi 4) 2 + (yi+5) 2 ] = 20
x 1 2 + 4x 1 +4+;y 1 2 6;y 1 + 9[x 1 2 8x 1 + 16 + ;y 1 2 +10;y 1 + 25] = 20
12*! 16^48 =
i.e. 3xj  4y l  12 =
The locus of (Xpjj) is 3x  Ay  12 =
Example 5.4: Find a point on xaxis which is equidistant from the points
(7,  6) and (3, 4) .
Solution:
Let P(xj, y 1 ) be the required point. Since P lies on xaxis, y^ = 0.
Given that A(7,  6) and B(3, 4) are equidistant from P.
i.e. PA = PB => PA 2 = PB 2
=> (xj  7) 2 + (0 + 6) 2 = (xj  3) 2 + (0  4) 2
134
=> xj 2  14xj + 49 + 36 = xj 2  6x l + 9 + 16
=> 8xj = 60 .. x 1 = 15/2
Thus the required point is [pr ,0
EXERCISE 5.1
(1) A point moves so that it is always at a distance of 6 units from the point
(1,  4). Find its locus.
(2) Find the equation of the locus of the point which are equidistant from
(1,4) and (2, 3).
(3) If the point P(5f  4, t + 1) lies on the line 7x  Ay + 1 = 0, find
(i) the value of t (ii) the coordinates of P.
(4) The distance of a point from the origin is five times its distance from
the yaxis. Find the equation of the locus.
(5) Show that the equation of the locus of a point which moves such that its
distance from the points (1, 2) and (0,1) are in the ratio 2 : 1 is 3x +
3y 2 + 2x+ 12jl=0.
(6) A point P moves such that P and the points (2, 3), (1, 5) are always
collinear. Show that the equation of the locus of P is 2x + y  7 = 0.
(7) A and B are two points (1, 0) and ( 2, 3). Find the equation of the
locus of a point such that (i) PA 2 + PB 2 = 10 (ii) PA = 4PB.
5.2 Straight lines
5.2.1 Introduction
A straight line is the simplest geometrical curve. Every straight line is
associated with an equation. To determine the equation of a straight line, two
conditions are required. We have derived the equation of a straight line in
different forms in the earlier classes. They are
(1) Slopeintercept form:
i.e. y = mx + c where 'm' is the slope of the straight line and V is the
y intercept.
(2) Pointslope form:
i.e. y — yj = m(x  x^) where c m' is the slope and (xp y^) is the given point.
(3) Two point form:
yy\ xxi
i.e = where (x,, v,) and (x , y ) are the two given points.
y2~yi X2X1 i J1 ' 2;2
135
(4) Intercept form:
x y
i.e. — + f = 1 where 'a' and 'b' axe x and y intercepts respectively.
In this section we shall derive and discuss other forms of equation of a
straight line.
5.2.2 Normal form:
Equation of a straight line in terms of the length of the perpendicular
p from the origin to the line and the angle a which the perpendicular
makes with jcaxis.
Let R and N be the points where the
straight line cuts the x and y axes
respectively.
Draw the perpendicular OL to RN.
Let OL = p and XOL = a.
Now OR and ON are the x and y
intercepts respectively.
Fig. 5. 3
x y
The equation of the straight line is t^t +7y\T =1 •••(!)
OR
From the right angled triangle OLR, sec a = 7^~ .'. OR = p sec a
ON
From the right angled triangle OLN, cosec a = sec (90  a) = j^r
.". ON =p coseca
Substituting the values of OR and ON in equation (1),
x y x cos a y sin a
we get, + = 1 i.e. —  — + —  — = 1
p sec a p cosec a P P
i.e. x cos a + y sin a = p is the required equation of the straight line.
5.2.3 Parametric form
Definition: If two variables, say x and y, are functions of a third variable,
say '9', then the functions expressing x and y in terms of 9 are called the
parametric representations of x and y. The variable 9 is called the parameter of
the function.
Equation of a straight line passing through the point (x v jj) and
making an angle 9 with xaxis. (parametric form)
136
Let Q (xj, yj) be the given point and P(x, y) be
any point on the required straight line. Assume
that PQ = r.
It is given that
PTR = 9. But PQM = PTR
.. PQM = 9
In the right angled triangle PQM,
...(1)
QM NR OR ON xx {
v ■■
i 1
Q
x %
M
A§
n
"1
fx o
N
K
Fig. 5. 4
cos9 =
PQ
x — xi
cos9
= r
n PM PR  MR
Similarly sin9 = p7j = ~
yy\
yy\
sin9
= r
(2)
From (1) and (2),
xx\ yy\
~z = — r~r = r which is the required equation.
cos
Any point on this line can be taken as (x\ + r cos 9, y\ + r sin 9) where r is
the algebraic distance. Here r is the parameter.
5.2.4 General form
The equation ax + by + c = will always represent a straight line.
Let (x,, y,), (x 2 , y 2 ) anc ^ ^ x 3' ^3) ^ e an y triree points on the locus
represented by the equation ax + by + c = 0. Then
ax, + by. + c = ••■(!)
ax, + by 2 + c = . . . (2)
ax 3 + by 3 + c = . . . (3)
(1) x (y 2  y 3 ) + (2) x (y 3  yj + (3) x(y l ~ y 2 ) gives
a [x[ (yi  yj) + x 2 O3  yi) + x 3 (yi  y 2 )] =
Since a * 0, Xj (y 2  y 3 ) + x 2 (y 3  yj + x 3 (y t  y 2 ) =
137
That is (xj, y^), (x 2 , y 2 ) anc ^ ( x 3' ^3) are collinear and hence they lie on a
straight line.
Thus the equation ax + by + c = represents a straight line.
5.2.5. Perpendicular distance from a point to a straight line
The length of the perpendicular from the point (xj,jj) to the line
ax\ + by\ + c
ax + by + c = is
(1)
V^7 2
Let the given line ax + by + c =
be represented by AB.
Let P(xj, jj) be the given point.
Draw PD perpendicular to AB. Note that
PDis
the required distance. x
Draw OM parallel to PD. Let OM = p
Assume that MOB = a.
Fig. 5. 5
From 5.2.2, the equation of the straight line AB is
x cosa + y sina  p = . . . (2)
Now equations (1) and (2) are representing the same straight line. Hence
their corresponding coefficients are proportional.
cos a sina  p
a ~ b ~ c
ap . pb
cosa = — , sina =  ~~
We know that
2 2
sin a + cos a = 1
2, 2 2 2
p b pa
2 2, 2,2
= 1 i.e. pa + p b = c
2^ 2 , , 2, 2 ■ 2 c
p (a +b ) = c i.e p = 2 ,2
a + b
Hence cosa =
= +
^[a
2 + b 2
, sina =
Va 2 + b 2
b
\]a 2 + b 2
138
Suppose OL = p', the equation of the straight line NR is
x cos a + y sin a  p' =
since P(Xp y^) is a point on NR
Xj cos a + y l sin a  /?' =
i.e. OL =p' =x l cos a + y^ sin a
From the figure, the required distance
PD = LM = OM  OL =p p'
= p  x, cosa  y, sin a
c x\ . a y\ . b ax\+byi+c
~ ^ja 2 + b 2 ~ \ja 2 + b 2 ^ja 2 + b 2 ~ \ja 2 + b 2
ax\ + by i + c
The required distance =
\[a
2 + b 2
Corollary:
The length of the perpendicular from the origin to ax + by + c = is
c
la 2 + b 2
c
4
Note: The general equation of the straight line is ax+by+c = i.e. y = Tx  ~
This is of the form y = mx + c.
a . coefficient of x
:. m =  B i.e. slope =  co . efficient of "
Example 5.5: Determine the equation of the straight line whose slope is 2 and
jintercept is 7.
Solution:
The slope  intercept form is y = mx + c Here m = 2, c = 7
.". The required equation of the straight line is y = 2x + 7
Example 5.6: Determine the equation of the straight line passing through
2
( 1,2) and having slope j
Solution:
The pointslope form isy y l = m(x  x^).
2
Here (xj, y^) = ( 1, 2) and m = ij
2
y2 = j(x+l) i.e. ly  14 = 2x + 2
2x  ly + 16 = is the equation of the straight line.
139
Example 5. 7:
Determine the equation of the straight line passing through the points
(1,2) and (3, 4).
Solution:
yyi xx\
The equation of a straight line passing through two points is =
y\yi x\X2
Here (x v y { ) = (1, 2) and (x 2 , y 2 ) = (3,  4).
y — 2 x — 1
Substituting the above, the required line is 9 + 4 = 1 — T
y  2 x  1 y  2 x  1
^ 6 = 2 ^ 3 = 1
=^> y2 = 3(xl)^>y2 = 3x + 3
=> 3x + y = 5 is the required equation of the straight line.
Example 5.8: Find the equation of the straight line passing through the point (1,
2) and making intercepts on the coordinate axes which are in the ratio 2:3.
Solution:
x y
The intercept form is — +f = 1 ■■■(I)
The intercepts are in the ratio 2:3 .'. a = 2k, b = 3k.
x y
(1) becomes jT + ~ti =i x  e  3x + 2y = 6k
Since (1, 2) lies on the above straight line, 3 + 4 = 6k i.e. 6k = 7
Hence the required equation of the straight line is 3x + 2y = 7
Example 5.9: Find the length of the perpendicular from (2,  3) to the line
2x  y + 9 =
Solution:
The perpendicular distance from (jcj, y^) to the straight line ax + by + c =
ax\ + by\ + c
is given by
V«
2 , j2
+ b
..The length of the perpendicular from (2, 3) to the straight line
2(2)  ( 3) + 9
2x  y + 9 = is
V(2) 2 + (D 2
16
■■^ units.
Example 5.10: Find the coordinates of the points on the straight line y = x + 1
which are at a distance of 5 units from the straight line 4x  3y + 20 =
140
Solution: Let (xj, y^) be a point on y = x + 1
:.y l =x 1 + l ...(1)
The length of the perpendicular from (xp y^) to the straight line
4x  3y + 20 = is
4xi  3yi + 20
, 4xi  3yi + 20
V4 2 + ( 3) 2
But the length of the perpendicular is given as 5.
^4xi  3yi + 20
5 ' = 5
:. 4xj  3y l + 20 = + 25
Considering the positive sign, 4xj  3y^ + 20 = 25
=> 4x l 3y l = 5 ... (2)
Considering the negative sign, 4xj  3yj + 20 =  25
=> 4x x  3y x =45 ...(3)
Solving (1) and (2), we get x 1 = 8, yj = 9
Solving (1) and (3), we get Xj=42, ^=41.
.". The coordinates of the required points are (8, 9) and ( 42,  41).
Example 5.11: Find the equation of the straight line, if the perpendicular from
the origin makes an angle of 120° with xaxis and the length of the
perpendicular from the origin is 6 units.
Solution:
The normal form of a straight line is x cosa + y sina = p
Here a = 120°, p = 6 :. x cos 120° + y sin 120° = 6
=> x i~2J +y { 2 J = 6 ^ x + y{3y= 12
=> xV3y+12 =
.". The required equation of the straight line is x  y[3 y + 12 =
Example 5.12: Find the points on yaxis whose perpendicular distance from the
straight line 4x  3y  12 = is 3.
Solution:
Any point on jaxis will have x coordinate as 0.
Let the point on yaxis be P(0, y^).
141
The given straight line is 4x  3j  12 = • • • (1)
The perpendicular distance from the point P to the given straight line is
3yi  12
3yi + 12
\J4 2 + ( 3) 2
5
But the perpendicular distance is 3.
i.e.
3yi + 12
5
= 3
=>
3y t + 12 = +15
3y 1 + 12 = 15
or
3y t + 12 =  15
3y 1= 3
or
3^ =27
)
1 =
1
or
>'!='
;
Thus the required points are (0, 1) and (0,  9).
EXERCISE 5.2
(1) Determine the equation of the straight line passing through the point
4
(1,2) and having slope 7
(2) Determine the equation of the line with slope 3 and jintercept 4.
(3) A straight line makes an angle of 45° with xaxis and passes through the
point (3,  3). Find its equation.
(4) Find the equation of the straight line joining the points (3, 6) and
(2,5).
(5) Find the equation of the straight line passing through the point (2, 2)
and having intercepts whose sum is 9.
(6) Find the equation of the straight line whose intercept on the xaxis is
3 times its intercept on the jaxis and which passes through the point
(1,3).
(7) Find the equations of the medians of the triangle formed by the points
(2, 4), (4, 6) and (6,10).
(8) Find the length of the perpendicular from (3, 2) to the straight line
3x + 2y+ 1=0.
(9) The portion of a straight line between the axes is bisected at the point
( 3, 2). Find its equation.
(10) Find the equation of the diagonals of a quadrilateral whose vertices are
(1,2), (2,1), (3, 6) and (6, 8).
142
(11) Find the equation of the straight line, which cut off intercepts on the
axes whose sum and product are 1 and  6 respectively.
(12) Find the intercepts made by the line Ix + 3y  6 = on the coordinate
axis.
(13) What are the points on xaxis whose perpendicular distance from the
x y
straight line ^ + J = 1 is 4?
(14) Find the distance of the line 4x  y = from the point (4, 1) measured
along the straight line making an angle of 135° with the positive
direction of the xaxis.
5.3. Family of straight lines
In the previous section, we studied about a single straight line. In this
section we will discuss the profile about more than one straight line, which lie
on a plane.
5.3.1 Angle between two straight lines
Let L:y = m*x + c, and y
Z 2 : y = m^x + c 2 be the two
intersecting lines and assume that P be
the point of intersection of the two
straight lines which makes angle 9j and
9 2 with the positive direction of xaxis.
Then m l = tan9j and m 2 = tan9 2  Let 9
be the angle between the two straight
lines.
Fig. 5. 6
From the figure (5.6),
9j = 9 + 9 2
.'. 9 = 9,— 9^
tan9 = tan (9j  9 2 )
 m.2
tan9i  tan92 m\—m.2
1 + tan0 1 .tan0 2 \+m\m2
mi ~ m 2
Note that f" ; is either positive or negative. As convention we
1 + m\ mi v b
consider the acute angle as the angle between any two straight lines and hence
we consider only the positive value (absolute value) of tan9.
Hence
tan9 =
m\ — ni2
1 + m\ m.2
9 = tan
m\ — m.2
1 + m\ m.2
143
Corollary (1) : If the two straight lines are parallel, then their slopes are equal.
Proof:
Since the two straight lines are parallel, 9 = 0. .'. tan 9 =
m.\  m.2
=> "T~ =0 => m. m~ =
1 + mi m.2 12
i.e. m, = niy
:. If the straight lines are parallel, then the slopes are equal.
Note : If the slopes are equal, then the straight lines are parallel.
Corollary (2): If the two straight lines are perpendicular then the product of
their slopes is 1.
Proof:
Since the two straight lines are perpendicular, 9 = 90°.
m\ — m.2
:. tanO = tan90° = oo => , = co
1 + m\ m2
This is possible only if the denominator is zero.
i.e. 1 + m, m 2 = i.e. m, m 2 =  1
.". If the two straight lines are perpendicular then the product of their
slopes is  1 .
Note (1): If the product of the slopes is  1, then the straight lines are
perpendicular.
(2): Corollary (2) is applicable only if both the slopes m^ and m 2 are
finite. It fails when the straight lines are coordinate axes or parallel
to axes.
Corollary (3): If the straight lines are parallel, then the coefficients of x and y
are proportional in their equations. In particular, the equations of two parallel
straight lines differ only by the constant term.
Proof:
Let the straight lines a^x + b^y + Cj = and a 2 x + b 2 y + c 2 = be parallel.
a\
Slope of a,x + b,y + c, = is m, =  ~r~ ; Slope of a 2 x + b 2 y + c 2 = is
(22
m 2=V 2
Since the straight lines are parallel, m l = m 2 .
a\ a.2 a\ b\
Le ' ~ b\ ~ ~ b2 ^ ci2 b2
144
i.e. coefficients of x and y are proportional
Let tt, = h = x ^
:. a 2 = a,X, b 2 = b, X
The second equation a 2 x + b 7 y + c 2 = can be written as
Xa , x + X b , y + c 2 =
c 2 . c 2
i.e. a,x + b,y + t~ =0 i.e. a,x + b,y + k = where k = ~r
i.e. If a^x + b^y + c^ = is a straight line then a line parallel to it is
a,x + b,y + k =
:. Equations of parallel straight lines differ by the constant term.
Note (1): In the previous section, we established a formula to find the
distance between the origin and the straight line. i.e. distance =
"V^
2 + b 2
We can find out the distance between two parallel straight lines
I c\ —ci I
ax + by + Cj = and ax + by + c 2 = by using the formula d = , .
~\ja + b
This is obtained by using the above result. Note that, we took I Cj  c 2 I since
Cy > c, or c, > c 2
Note (2): To apply the above formula, write the equations of the parallel
straight lines in the standard form ax + by + c l = and ax + by + c 2 = 0.
Corollary (4): The equation of the straight line perpendicular to the straight
line ax + by + c = is of the form bx  ay + k = for some k.
Proof:
Let the straight lines ax + by + c = and a^x + b^y + c^ = be
perpendicular.
a
Slope of ax + by + c = is m^ =  t
a\
Slope a,x + b.y + c, = is m 2 = — ~r~
Since the straight lines are perpendicular, m^m 2 = \
i.e. T7 T =1 i.e. aa, = bb.
l \
145
Cl\
i.e.
= X (say)
.". a, = bX and b,=aX
The second equation ajX + ^y + Cj =0 can be written as bXx  aXy + c^ =
i.e.
i.e.
bx ay + ~r =
bx ay + k = where fc =
£1
A straight line perpendicular to ax + by + c = is given by bx  ay + k =
for some k.
Note: To find the point of intersection of two straight lines, solve the
simultaneous equations of the straight lines.
5.3.3 The condition for the three straight lines to be concurrent
Let the three straight lines be given by
ayX + b^y + Cj = ... (1)
a 2 x + b 2 y + c 2 = . . . (2)
a^x + b^y + c 3 = . . . (3)
If the three straight lines are concurrent, then the point of intersection of
any two straight lines lies on the third straight line.
Solving the equation (1) and (2), the coordinates of the point of
intersection is
b\ c 2 b 2 c\ c\a 2 c 2 a\
x — v —
a\b 2  a 2 b\ a\b 2  a 2 b\
substituting the values of x and y in the equation (3)
b\ c 2  b 2 c\
3\a\b 2  a 2 b[) ^\a\b 2 — a 2 b\
c\a 2 c 2 a\
+ c 3 =
i.e. <au Q ) \ c 2 ~ b 2 c,) +bJc,a 2  c 2 aS) + cAaJjy ~ <*J>\) =
i.e. a^b^c^  b^c 2 )  b^(a 2 c^
a\ b\ c\
3 c 2 ) + c^(a 2 b^  a^b 2 ) =
i.e.
a 2 b 2 c 2
aj, b 3 c 3
concurrent.
= is the condition for the three straight lines to be
146
5.3.4 Equation of a straight line passing through the
intersection of the two given straight lines
Let a ] x + b l y + c^ =
a 2 x + b 2 y + c 2 =
(1)
(2)
(3)
be the equations of the two given straight lines.
Consider the equation a^x + b^y + c^ + X (a 2 x + b 2 y + c 2 ) =
where X is a constant
Equation (3) is of degree one in x and y and therefore (refer 5.2.4) it
represents a straight line. Let (x { , y^) be the point of intersection of (1) and (2)
.". a,x, + b^y, + c, = and a 2 x, + b 2 y, + c 2 =
.'. a,x, + b,y, + c, + X (a 2 x. + b 2 y, + c 2 ) =
.". Value of (xp y^) satisfies equation (3) also.
Hence a^x + b^y + c^ + X (a 2 x + b 2 y + c 2 ) = represents a straight line
passing through the intersection of the straight lines a^x + b { y + c^ = and
a,x + b 2 y + c 2 =
Example 5.13: Find the angle between the straight lines 3x  2y + 9 = and
2x + y  9 = 0.
Solution:
3 9'
■■y=2 x+ 2
Slope of the straight line 3x  2y + 9 = is m.y =j
Slope of the straight line 2x + y  9 = is m 2 =  2 [ v y =  2x + 9]
Suppose '9' is the angle between the given lines, then
1
9 = tan
= tan
= tan
1 + m\ m.2
2 + 1
l+(2)
= tan
7
2
26
= tan
Example 5.14: Show that the straight lines 2x + y  9 = and 2x + y  10 =
are parallel.
147
Solution:
Slope of the straight line 2x + y9 = 0ism^ = 2
Slope of the straight line 2x + y  10 = is m 2 =  2 :. m l = m 2
.'. The given straight lines are parallel.
Example 5.15: Show that the two straight lines whose equations are
x + 2y + 5 = and 2x + Ay  5 = are parallel.
Solution:
The two given equations are
x + 2y + 5 =
2x + Ay  5 =
1 2
The coefficients of x and y are proportional since ~ = 7 and therefore they
are parallel.
Note : This can also be done by writing the equation(2) as x + 2y  5/2 =
Now the two equations differ by constant alone. .". They are parallel.
Example 5.16: Find the distance between the parallel lines 2x + 3y  6=0 and
2x + 3y + 1 = 0.
Solution:
C1C2
(1)
(2)
The distance between the parallel lines is
Here Cj =  6, c 2 = 7, a = 2, b = 3
V
a 2 + b 2
The required distance is
67
V2 2 + 3 2
=
13
Vl3
= ^/l3 units.
Example 5.17: Show that the straight lines 2x + 3y  9 = and 3x  2y + 10 =
are at right angles.
Solution:
Slope of the straight line
Slope of the straight line
2x + 3y  9 = is Wj =  ^
3
3x  2>> + 10 = is m 2 = 7
2 3
. . m,m 2 =  7 7 =  1
The two straight lines are at right angles.
148
Example 5.18: Find the equation of the straight line parallel to 3x + 2y = 9 and
which passes through the point (3,  3).
Solution:
The straight line parallel to 3x + 2y  9 = is of the form
3x + 2y + k = ... (1)
The point (3,  3) satisfies the equation (1)
Hence 9  6 + k = i.e. k =  3
.". 3x + 2y  3 = is the equation of the required straight line.
Example 5. 19: Find the equation of the straight line perpendicular to the
straight line 3x + Ay + 28 = and passing through the point ( 1, 4).
Solution:
The equation of any straight line perpendicular to 3x + Ay + 28 = is of the
form
Ax  3y + k =
The point ( 1,4) lies on the straight line Ax  3y + k =
.. 412 + fe = => k = 16
.". The equation of the required straight line is Ax  3y + 16 =
Example 5. 20: Show that the triangle formed by straight lines
Ax  3y  18 = 0, 3x  Ay + 16 = and x + y2 = 0is isosceles.
Solution:
Slope of the straight line Ax  3y ■
18 = is m,y = t
Slope of the straight line 3x  Ay + 16 = is m 2 = T
Slope of the straight line x + >>2 = 0ism 3 =  1
Let 'a' be the angle between the straight lines Ax
3x Ay +16 =
tri[ mi
Using the formula, 9 = tan
1 + m\ OT2
we get
a = tan
= tan
= tan
24
3y  18 = and
4 3
169
3"4
, 4 3
1 + 34
= tan 1
12
2
149
Let '(3' be the angle between the straight lines 3x 4y+l6 = and x + y  2 =
7/4
(3 = tan" 1
!♦■
i+l^" 1 )
= tan
■l
1/4
= tan" 1 (7)
Let 'y' be the angle between the straight lines x+y 2 = and 4x3y  18=
.". y = tan
l + (Dl3
= tan
= tan" 1 (7)
Therefore (3 = y •'• The triangle is isosceles.
Example 5.2i:Find the point of intersection of the straight lines
5x + Ay  13 = and 3x + y  5 =
Solution:
To find the point of intersection, solve the given equations.
Let (xj, jj) be the point of intersection. Then (jci, y\) lies on both the
straight lines.
.. 5x 1 +4y 1 = 13 ...(1)
3^+^ = 5 ...(2)
(2)x4 ^ 12x^4^=20 ...(3)
(l)(3) =$ lx x =7 .. x x = \
Substituting x^ = 1 in equation (1), we get 5 + Ay^ = 13
Ay x =8 .". y, = 2
The point of intersection is (1, 2).
Example 5.22: Find the equation of the straight line passing through the
intersection of the straight lines 2x + y = 8 and 3x  y = 2 and through the point
(2,3)
Solution:
The equation of the straight line passing through the intersection of the
given lines is
2x + y8 + X(3xy2) = ...(1)
(2,  3) lies on the equation (1) and hence A3  8 + X (6 + 3 2) =
150
.. X = 1
.. (1) => 2x + y8 + 3xy2 = => 5jc10 =
x = 2 is the equation of the required straight line.
Example 5.23: Find the equation of the straight line passing through the
intersection of the straight lines 2x + y = 8 and 3x  2y + 7 = and parallel to Ax
+ yll=0
Solution:
Let (xp jj) be the point of intersection of the given straight lines
2xy + y l = 8
...(1)
3xj  2jj =  7
...(2)
(l)x2 =>
4xj + 2y l = 16
...(3)
(2) + (3) =>
9
■'■ x \ ~ 1
38 (9 38^
y 1= 7 .. (x 1 ,y 1 ) = ( J , 7 J
The straight line parallel to 4x + y 
1 1 = is of the form Ax + y + k =
But it passes through \j ,y 1
36 38
■y +^y" +fc =
• k  7 4
74
4jc + j  y =0
28x + ly  1A = is the equation of the required straight line.
Example 5.24:
Find the equation of the straight line which passes through the intersection
of the straight lines 5x  6y = 1 and 3x + 2y + 5 = and is perpendicular to the
straight line 3x  5y + 1 1 =
Solution:
The straight line passing through the intersection of the given straight lines is
5x6yl+X(3x + 2y + 5) = ...(1)
(5 + 3X)x + ( 6 + 2X)y + ( 1 + 5X) =
This straight line is perpendicular to 3x  5y + 1 1 =
Product of the slopes of the perpendicular straight lines is 1 i.e. ?Mj m 2 = 1
5 + 3^ ^ (?>,
6 + 2XJ \5;
15 + 9?. = 30+m ..A. = 45
151
(1) => 5x  6y  1 + 45 (3x + 2y + 5) = i.e. UOx + My + 224 =
i.e. 5x + 3y + 8 = is the equation of the required straight line.
Example 5.25: Show that the straight lines 3x + 4y = 13; 2x  ly + 1=0 and
5x  y = 14 are concurrent.
Solution:
Let (xj, jj) be the point of intersection of the first two straight lines
3*1 + %i
2xi
" 7 ?l
(l)x7
(2)x4
(3) + (4)
(1)
2lx l + 28y l =
8xj  28^ =
29xj =
13
...(1)
1
...(2)
91
...(3)
4
...(4)
87 => jcj = 3
13 => y, =1
=> 9 + Ay x
The point of intersection of the first two straight lines is (3, 1).
Substitute this value in the equation 5x  y = 14
L.H.S. = 5xy
= 15  1 = 14 = R.H.S.
i.e. The point (3, 1) satisfies the third equation.
Hence the three straight lines are concurrent.
Example 5.26: Find the coordinates of orthocentre of the triangle formed by
the straight lines
j C  } ;5 = 0, 2xy8 = 0and3jc)'9 =
Solution:
Let the equations of sides AB, BC
and CA of a AABC be represented by
X y5 = ...(1)
2xy8 = ...(2)
3xy9 = ...(3)
Solving (1) and (3), we get A as (2,  3)
B
The equation of the straight line BC is 2x
perpendicular to it is of the form
x + 2y + k =
D 2xyS =
Fig. 5. 7
8 = 0. The straight line
...(4)
152
A(2,  3) satisfies the equation (4) .. 2  6 + k = =^> A; = 4
The equation of AD is x + 2y =  4 . . . (5)
Solving the equations (1) and (2), we get B as (3,  2)
The straight line perpendicular to 3x  y  9 = is of the form
x + 3y + k =
But B(3,  2) lies on this straight line :. 36 + k = => k = 3
.". The equation of BE is x + 3y =  3 . . . (6)
Solving (5) and (6), we get the orthocentre O as ( 6, 1).
Example 5.27: For what values of 'a', the three straight lines 3x + y + 2 = 0,
2x  y + 3 = and x + ay  3 = are concurrent?
Solution:
Let (xp yj) be the point of concurrency. This point satisfies the first two
equations.
.. 3x 1 +y 1 + 2 = ...(1)
2xjyj + 3 = ...(2)
Solving (1) and (2) we get ( 1, 1) as the point of intersection. Since it is a
point of concurrency, it lies on x + ay  3 = 0.
..  1 + a  3 =
i.e. a = 4
EXERCISE 5.3
(1) Find the angle between the straight lines 2x + y = 4 and x + 3y = 5
(2) Show that the straight lines 2x + y = 5 and x  2y = 4 are at right angles.
(3) Find the equation of the straight line passing through the point (1,  2)
and parallel to the straight line 3x + 2y  7 =
(4) Find the equation of the straight line passing through the point (2, 1)
and perpendicular to the straight line x + y = 9
(5) Find the point of intersection of the straight lines 5x + 4y  13 =
and 3x + y  5 =
(6) If the two straight lines 2x  3y + 9 = 0, 6x + ky + 4 = are parallel,
find k
(7) Find the distance between the parallel lines
2x + y  9 = and 4x + 2y + 7 =
(8) Find the values of p for which the straight lines 8px + (2  3p) y + 1 =
and px + 8y  7 = are perpendicular to each other.
153
(9
(10
(11
(12
(13
(14
(15
(16
(17
(18
(19
(20
(21
(22
Find the equation of the straight line which passes through the
intersection of the straight lines 2x + y = 8 and 3x  2y + 7 = and is
parallel to the straight line Ax + y  11 =
Find the equation of the straight line passing through intersection of the
straight lines 5x  6y = 1 and 3x + 2y + 5 = and perpendicular to the
straight line 3x  5y + 1 1 =
Find the equation of the straight line joining (A,  3) and the
intersection of the straight lines 2x  y + 7 = and x + y  1 =
Find the equation of the straight line joining the point of the intersection
of the straight lines 3x + 2y + 1 = and x + y = 3 to the point of
intersection of the straight lines y — x = 1 and 2x + y + 2 =
Show that the angle between 3x + 2y = and Ax  y = is equal to the
angle between 2x + y = and 9x + 32y = 41
Show that the triangle whose sides are y = 2x + 7, x  3y  6 = and
x + 2y = 8 is right angled. Find its other angles.
Show that the straight lines 3x + y + A = 0, 3x + Ay  15 = and
24x  ly  3 = form an isosceles triangle.
Show that the straight lines 3x + 4>>=13; 2x  ly + 1 = and 5x  y = 14
are concurrent.
Find 'a' so that the straight lines x  6y + a = 0, 2x + 3y + A = and
x + Ay + 1 = may be concurrent.
Find the value of 'a' for which the straight lines
x + yA = 0,3x + 2 = and x  y + 3a = are concurrent.
Find the coordinates of the orthocentre of the triangle whose vertices
are the points ( 2,  1), (6,  1) and (2, 5)
If ax + by + c = 0, bx + cy + a = and ex + ay + b = are concurrent,
show that a 3 + b 3 + c 3 = 3abc
Find the coordinates of the orthocentre of the triangle formed by the
straight lines x + y  I =0, x + 2y  4 = and x + 3y  9 =
The equation of the sides of a triangle are x + 2y = 0, Ax + 3y = 5 and
3x + y = 0. Find the coordinates of the orthocentre of the triangle.
5.4 Pair of straight lines
5.4.1 Combined equation of the pair of straight lines
We know that any equation of first degree in x and y represents a straight line.
Let l^x + m^y + n^ = and l 2 x + m 2 y + n 2 = be the individual equations of any
two straight lines. Then their combined equation is
(Lx + m.y + «,) (Lx + m^y + « 2 ) =
154
2 2
ZjZ 2 x + (/jm 2 + /2 OT i) *y + m^rn^y +(1^2 + / 2 M i) x + (tn^ + m^n^y + n^x^ =
Hence the equation of a pair of straight lines may be taken in the form
2 2
ax + 2hxy + by + 2gx + 2fy + c = 0, where a, b, c,f, g, h are constants.
5.4.2 Pair of straight lines passing through the origin
2 2
The homogeneous equation ax + 2hxy + by = of second degree in
x and y represents a pair of straight lines passing through the origin.
2 2
Considering ax + 2hxy + by = as a quadratic equation in x, we get
/ 2 2 2
2hy ±\J4 h y Aaby
la
2h± 2\jh 2  ab
2a
y
h ± \lh 2  ab
y
= (h±^h 2 ab)
y
i.e. ax+\h+ \]h  ab) y = and ax + \h 'Sjh  ab) y = are the
2 2
two straight lines, each passing through the origin. Hence ax + 2hxy + by =
represents a pair of straight lines intersecting at the origin.
2
Note : The straight lines are (1) real and distinct if h > ab
2
(2) coincident if h = ab
2
(3) imaginary if h <ab
Sum and product of the slopes of pair of straight lines
2 2
The homogeneous equation ax + 2hxy + by = of second degree in x and
y represents a pair of straight lines passing through the origin.
Let y = m^x and y = m^x be the two straight lines passing through the
origin. Therefore the combined equation is (y — m^x) (y  m^x) =
2 2
=> mj/fiji (m^+m^xy + y =0
This equation also represents a pair of straight lines passing through the
origin.
Equating the coefficients of like terms in the above equations, we get
m\m.2 (m\ + mj) \
a ~ 2h ~ b
a a
:. m,y m 2 = r ; i.e. Product of the slopes = t
2h . „ r , , 2h
m,y + m 2 = —~i~ i.e. Sum of the slopes = ~r
155
5.4.3 Angle between pair of straight lines passing through the
origin
The equation of the pair of straight lines passing through the origin is
2 2
ax + 2hxy + by =
1h a
m.+m2 = —~iT and m. m 2 = r
Let '9' be the angle between the pair of straight lines.
m\  m.2
(1)
tanG =
tan 9 =
1 + m\ m.2
\j(m[ + m.2) 4m[tn2
1 + otj m.2
\ \hf 4a
b 2 ~ b
a
l+r
4h z  4ab
a + 1
tan 9 =
+ 2\jh 2  ab
a + b
9 = tan" 1
±2\jh 2
a + 1
 ab
»
i.e. 9 = tan
■d
2\h^ab
a + 1
It is conventional to take 9 to be acute.
Corollary (1):
If '9' is the angle between the pair of straight lines
2 2
ax + 2hxy + by + 2gx + 2fy + c =
then 9 = tan
2 \jh 2  ab
a + b
It is same as the angle between the pair of straight lines
2 2
ax + 2hxy + by = passing through the origin.
Corollary (2): If the straight lines are parallel, then h =ab
[since 9 = 0°, tan9 = 0]
Corollary (3): If the straight lines are perpendicular then
coeff. of x 2 + coeff. of y 2 = [since 9 = 90°, tan9 = oo]
156
The condition for a general second degree equation
2 2
ax + 2hxy + by + 2gx + 2fy + c = to represent a pair of straight lines
is abc + 2fgh  af  bg 2 ch 2 =
2 2
Assume that ax + 2hxy + by + 2gx + 2fy + c = ■■■(!)
represents a pair of straight lines. Treating this equation as a quadratic in x,
;an be written as ax +
By solving for i,we get
2 2
this can be written as ax + 2(hy + g)x + (by + 2fy + c) =
x =
(hy + g)± yjjhy + R) 2  a(by 2 + 2fy + c)
ax ■
+ hy + g = ± yj (hy + g) 2  a(by 2 + 2fy + c)
= ± \l(h 2  ab)y 2 + 2(gh  af)y + (g 2  ac)
Now in order that each of these equations may be of the first degree in
x and y, the expression in the R.H.S should be a perfect square. This is possible
only if the discriminant of this quadratic in 'y' under the radical or within the
root is zero.
•'• (h  ab) (g  ac) = (gh  af)
Simplifying this we get abc + 2fgh  af'  bg  ch =0 which is the
required condition.
2 2
Example 5.28: Find the angle between the straight lines x + Axy + 3y =0
Solution:
Here a = l,2h = 4,b = 3
If '9' is the angle between the given straight lines, then
9 = tan
d
2W  ab
a + 1
= tan
2^/4^3"
= tan
2 2
Example 5.29: The slope of one of the straight lines of ax + 2hxy + by = is
2
thrice that of the other, show that 3h = Aab
Solution:
Let 'mj' and 'm 2 ' be the slopes of pair of straight lines.
2h a
Then m, + m 2 = — ~r , m, m 2 = r
It is given that m 2 = 3m j
2h
m [ + 3m l = ~r =>
h_
2b
157
2_a _^,fz*¥_
But m l . 3otj = t =^> 3m 1 = » ~^ D \2b )
3h 2 a
=> 3/j 2 = 4a6
2 2
Example 5.30: Show that x  y +x3y2 = represents a pair of straight
lines. Also find the angle between them.
Solution:
The given equation is
x 2 y 2 + x3y2 = ...(1)
2 2
Comparing this with ax + 2hxy + by + 2gx + 2fy + c = we get a = 1,
13
h = 0,b = l,g = j,f=j , c =  2. Condition for the given equation to
2 2 2
represent a pair of straight lines is abc+2fghaf bg ch =
abc+2fghafbg 2 ch 2 = (l)(l)(2)+2( 1) (£) (0)(l) (f)  1) (£)(2) (0)
9 1 89+1
 2 "4 +4  4
=
Hence the given equation represents a pair of straight lines.
Since a + &=ll=0, the angle between the straight lines is 90°.
2 2
Example 5.31: Show that the equation 3x + Ixy + 2y + 5x + 5y + 2 =
represents a pair of straight lines and also find the separate equation of the
straight lines.
Solution:
2 2
Comparing the given equation with ax + 2hxy + by + 2gx + 2fy + c = 0,
we get
7 5 5
a = 3, b = 2, h = j , g = j , f = ^ , c = 2. The condition for the given
equation to represent a pair of straight lines is abc + 2fgh  aj bg  ch =0
abc+2fghaj 2 bg 2 ch 2 = (3) (2) (2) + 2(f) (f) (J)  3 (f) 2 (f) 2^
175 75 50 98 _
I2+4 _ 4 _ 4 _ 4 _
158
Hence the given equation represents a pair of straight lines.
Now, factorising the second degree terms
we get 3x 2 + Ixy + 2y 2 = (x + 2y) (3x + y)
Let 3x 2 + Ixy + 2y 2 + 5x + 5y + 2 = (x + 2y + I) (3x + y + m)
Comparing the coefficient of x, 31 + m = 5 ;
Comparing the coefficient of y, 1 + 2m = 5
Solving these two equations, we get I = l,m = 2
:. The separate equations are x + 2y + 1 = and 3x + y + 2 =
2 2
Example 5.32: Show that the equation 4x + 4xy + y  6x  3y  4 =
represents a pair of parallel lines and find the distance between them.
Solution:
2 2
The given equation is 4x + 4xy + y 6x3y 4 =
Here a = 4, h = 2, b=\ ; ab  h 2 = 4(1) 2 2 = 44 =
.'. The given equation represents a pair of parallel straight lines.
Now 4x 2 + 4xy + y 2 = (2x + y) 2
:. 4x 2 + 4xy + y 2  6x  3y  4 = (2x + y + I) (2x + y + m)
Comparing the coefficient of x, 21 + 2m =  6 i.e. l + m = 3 ■■■(!)
Comparing the constant term, lm = 4 . . . (2)
/ + ^yJ =3 ^ I 2 + 314 =
i.e. (I + 4) (I  1) = => I =  4, 1
Now Im = 4 => to = 1, — 4
.". The separate equations are 2x + y  4 = and 2x + y + 1 =
I ci  C2 I 41
The distance between them is
= ^/5 units
yja 2 + b 2 \J2 2 + l 2
Example 5.33: Find the combined equation of the straight lines whose separate
equations are x + 2y  3 = and 3x  y + 4 =
Solution:
The combined equation of the given straight lines is
(x + 2y3)(3xy + 4) =
i.e. 3x 2 + 6xy  9x  xy  2y 2 + 3y + 4x + 8y  12 =
2 2
i.e. 3x + 5xy — 2y  5x + lly  12 = is the required combined
equation.
159
EXERCISE 5.4
2 2
(1) If the equation ax + 3xy  2y  5x + 5y + c = represents a pair of
perpendicular straight lines, find a and c.
(2) Find the angle between the pair of straight lines given by
(a 2  3b 2 ) x 2 + Sab xy + fb 2  3a 2 )y 2 =
(3) Show that if one of the angles between pair of straight lines
ax 2 + 2hxy + by 2 =0 is 60° then (a + 3b) (3a + b) = Ah 2
(4) Show that 9x 2 + 2Axy + I6y 2 + 2lx + 2%y + 6 = represents a pair of
parallel straight lines and find the distance between them.
2 2
(5) The slope of one of the straight lines ax + 2hxy + by = is twice that
9
of the other, show that 8h = 9ab.
(6) Find the combined equation of the straight lines through the origin, one
of which is parallel to and the other is perpendicular to the straight line
2x + y + 1 =
(7) Find the combined equation of the straight lines whose separate
equations are x + 2y  3 = and 3x + y + 5 =
2 2
(8) Find k such that the equation 12* + Ixy  \2y x + ly + k =
represents a pair of straight lines. Find the separate equations of the
straight lines and also the angle between them.
2 2
(9) If the equation 12x  lOxy + 2y + \Ax  5y + c = represents a pair of
straight lines, find the value of c. Find the separate equations of the
straight lines and also the angle between them.
2 2
(10) For what value of k does 12* + Ixy + ky + I3x y + 3 = represents a
pair of straight lines? Also write the separate equations.
(11) Show that 3x 2 + lOxy + 8y 2 + \Ax + 22y + 15 = represents a pair of
_, (2^
straight lines and the angle between them is tan I tt
5.5 Circle
Definition: A circle is the locus of a point which moves in such a way that
its distance from a fixed point is always constant. The fixed point is called the
centre of the circle and the constant distance is called the radius of the circle.
160
()
0P (x, y)
Fig. 5.8
5.5.1 The equation of a circle when the centre and radius are given
Let C(h, k) be the centre and r be the
radius of the circle. Let P(x, y) be any point
on the circle
CP = r => CP 2 = r 2 => (x  h) 2 + (yk) 2 = r 2
is the required equation of the circle.
Note:
If the centre of the circle is at the origin,
i.e. (h, k) = (0, 0) then the equation of the
2 2 2
circle is x +y = r .
5.5.2 The equation of a circle if the end points of a diameter are
given
Let A(xj, Vj) and B(x 2 , y 2 ) be the end
points of a diameter. Let P(x, y) be any point
on the circle.
The angle in a semi circle is a right angle.
.". PA is perpendicular to PB
.. (Slope of PA) (Slope of PB) =  1
fyyi] (yyi\
x
o
P (X, V)
(A;. V 2 )
= 1
Fig. 5.9.
\x  x\) \x — X2J
(y y{)(y y 2 ) = (* *i) (x  x 2 )
:. (x  Xj) (x  x 2 ) + (y  Vj) (y  v 2 ) = is the required equation of the
circle.
5.5.3 The general equation of the circle is x 2 + y 2 + 2gx + 2fy + c =
2 2
Consider the equation x + y + 2gx + 2fy + c=
This can be written as
x 2 + 2gx + g 2 + y 2 + 2fy +J 2 = g 2 +J 2 c
(x + g) 2 + (y +f) 2 = (^g 2 +f 2  c ) 2
[x  ( g)f +[y (f)f = (V# 2 +/ 2 c) 2
2 2 2
This is of the form (xh) +(yk) = r
:. The considered equation represents a circle with centre ( g, —J) and
radius ^g 2 +f 2 ■
2 2
.'. The general equation of the circle is x +y + 2gx + 2fy + c =
161
2 2
Note : The general second degree equation ax +by + 2hxy + 2gx + 2fy + c =
2 2
represents a circle if (Y)a = b i.e. coefficient of x = coefficient of y
(2) h = i.e. no xy term
5.5.4 Parametric form
Consider a circle with radius r and
centre at the origin. Let P(x, y) be any point
on the circle. Assume that OP makes an angle
9 with the positive direction of xaxis. Draw
the perpendicular PM to the xaxis.
X V
From the figure (5.10), ~ = cos9, ~; = sin9.
Fig. 5.10
Here x and y are the coordinates of any point on the circle. Note that
these two coordinates depend on 9.
The value of r is fixed. The equations x = r cos9, y = r sin9 are called
2 2 2
the parametric equations of the circle x + y = r . Here '9' is called the
parameter and < 9 < 2n
Another parametric form:
We know that
sin 9 =
2 tany
i 2~e
1 + tan 2
cos9 =
1  tan z 2
i 2©
1 + tan 2
Let t = tan y
If < 9 < 2n then  oo < t < oo
x = r cos 9
ril  n
x= ^~
i + r
Thus x =
rd  a
l + t 2
2rt
2 2 2
equation of the circle x + y = r
r(\  t 2 ) 2rt
y = r sin9
2rt
y
l + r
co < f < co is another parametric
Clearly x= ,
■ y =
i + t
2 2 2
2 satisfy the equation x + y = r
Example 5.34: Find the equation of the circle if the centre and radius are
(2,  3) and 4 respectively.
162
Solution:
2 2 2
The equation of the circle is (xh) +(yk) =r
Here (h, k) = (2,  3) and r = 4 .. (x  2) 2 + (y + 3) 2 = 4 2
2 2
i.e. x + y  4x + 6y  3 = is the required equation of the circle.
Example 5.35: Find the equation of the circle if (2,  3) and (3, 1) are the
extremities of a diameter.
Solution:
The equation of the circle is (x  Xj) (x — x 2 ) + (y — y^) (y  y 2 ) =
Here (x v y { ) = (2,  3) and (x 2 , y 2 ) = (3, 1)
.". (x2)(x3) + (y + 3)(yl) =
x 2  5x + 6 + y 2 + 2y  3 =
2 2
.". The required equation isx +y 5x + 2y + 3 =
2 2
Example 5.36: Find the centre and radius of the circle x +>> +2x4y + 3 =
Solution:
2 2
The general equation of the circle is x + y + 2gx + Ify + c =
Here 2g = 2, 2/=  4, c = 3
.". centre is ( g, /) = (— 1, 2)
radius is *\jg +fc= a/1 +43 = \/2 units.
2 2
Example 5.37: Find the centre and radius of the circle 3x +3y 2x+6y 6 =
Solution:
2 2
The given equation is 3x +3y  2x + 6y  6 =
2 2 ^
Rewriting the above, x +>> tx + 2>>2 =
2 2
Comparing this with the general equation x + y + 2gx + 2fy + c =
We get 2g = f,2/=2,c = 2
.. centre is (g,f) = (j,  1
radius is "\/g +f c =a/q+1+2 = ^ units.
Example 5.38: If (4, 1) is one extremity of a diameter of the circle
2 2
x + y  2x + 6y  15 = 0, find the other extremity.
163
Solution:
Comparing x + y  2x + 6y  15 = with
the general equation of the circle, A(4, I ) [ * ) B (*l < >i )
we get 2g =  2 2f= 6
.. centre is C ( g, f) = (1,  3) Fig. 5.11
Let B(x v yj) be the other extremity and
A be (4, 1)
C is the mid point of AB
x\ +4 yi + 1
— 2~ = 1, ^— =3 => x [= 2, y 1= l
:. The other extremity is ( 2,  7)
Example 5.39: Find the equation the circle passing through the points
(0,1), (2,3) and (2, 5).
Solution:
The general
2
equation of the circle is x
+ /
'■ + 2gx + 2fy + c =
The points (0, 1), (2, 3) and ( 2, 5) lie on the circle
••• 2/+c
=
 1
...(1)
4^ + 6/+ c
=
13
...(2)

4g + 10/+ c
=
29
...(3)
(l)(2)
=>
4g4f
=
12
8+f
=
3
...(4)
(2)  (3)
=>
*g¥
=
16
2g~f
=
4
...(5)
(4) + (5)
=>
3g
=
1 => g = i
(4)
=>
/
=
31
10
3
(1)
=>
c
=
17
3
• 2,2,
.. x +y +
2@x + 2(
IG\ 17
:. 3x 2 + 3y 2
+ 2jc  20y +
17 = is the required equation.
Example 5.40: Find the equation of the circle passing through the points
(0, 1), (2, 3) and having the centre on the line x  2y + 3 =
164
Solution:
The general equation of the circle is
=1
...(1)
=13
...(2)
:g + 2f=3
...(3)
AgAf =12
i.e. g+f= 3
•••(4)
3/= 6
•••/= 2
s = i
c = 3
x 2 + y 2 + 2gx + 2fy + c =
(0, 1) lies on the circle .". 2f+ c
(2, 3) lies on the circle .". Ag + 6f+ c
The centre ( g, —J) lies on x  2y + 3 = ;
(1)  (2) =>
(3) + (4) =>
(3) =>
(1) =>
2 2
.". The required equation is x +y 2x4y + 3 =
Example 5.41:
Find the values of a and b if the equation
2 2
(a  4)x + by + (b  3)xy + Ax + Ay  1 = represents a circle.
Solution:
2 2
The given equation is (a  A)x + by + (b  3)xy + Ax + Ay  1 =
(i) coefficient of xy = => b  3 = .'. b = 3
2 2 t
(ii) coefficient of x = coefficient ofj => a A = b
:. a = 7
Thus a = 7, b = 3
Example 5.42: Find the equation of the circle with centre (2,  3) and radius 3.
Show that it passes through the point (2, 0).
Solution:
If the centre is (h, k) and radius is r, then the equation of the circle is
(xh) 2 + (yk) 2 = r 2 .
Here (h, k) = (2,  3) and r = 3.
(x  2) 2 + (y + 3) 2 = 3 2
2 2
(x  2) + (y + 3) = 9 is the required equation of the circle.
Putting (2, 0) in the equation of the circle, we get
L.H.S. = (2  2) 2 + (0 + 3) 2 = + 9 = 9 = R.Hi
Hence the circle passes through (2, 0)
165
Example 5.43:
Find the equation of the circle with centre (1,  2) and passing through the
point (4, 1)
Solution:
Let C be (1, 2) and P be (4, 1)
1 P(4, 1 )
Radius r = CP = \j(l 4) 2 + (2  l) 2 = ^9 + 9 = a/T8
2 2 2 2
Thus the equation of the circle is (x  h) +(yk) =r =r
__ 2 Fig. 5.12
=> (x1) 2 + (v + 2) 2 = a/I8
2 2
i.e. x + >> 2x + 4>>13=0 is the required equation.
2 2
Example 5.44: Find the parametric equations of the circle x + y =16
Solution:
Here r = 16 => r = 4 . The parametric equations of the circle
2 2 2
x + y = r in parameter 9 are x = r cos9, y = r sin 9
2 2
.". The parametric equations of the given circle x + y = 16 are
x = 4 cos 9, y = 4 sin 9, < 9 < 2n
Example 5.45: Find the cartesian equation of the circle whose parametric
equations are
x = 2 cos 9, y = 2 sin 9, < 9 < 2%
Solution:
To find the caretsian equation of the circle, eliminate the parameter '9'
x v
from the given equations, cos 9 = x ; sin 9 = ~
cos 2 9 + sin 2 9 = 1 => ( J + f J =1
2 2
.". x + j = 4 is the required cartesian equation of the circle.
EXERCISE 5.5
(1) Find the centre and radius of the following circles:
(i) x 2 + y 2 = 1 (ii) x 2 + y 2  4x  6y  9 =
(iii) x 2 + y 2  8x  6y  24 = (iv) 3x 2 + 3y 2 + 4x  Ay  4 =
(v)(jc3)(jc5) + (y7)(yl) =
(2) For what values of a and b does the equation
2 2
(a  2)x + by + (b — 2)xy + Ax + Ay  1 = represents a circle? Write
down the resulting equation of the circle.
(3) Find the equation of the circle passing through the point (1, 2) and
having its centre at (2, 3).
166
(6)
(7)
(8)
(9)
(4) x + 2y = 7, 2x + y = 8 are two diameters of a circle with radius 5 units.
Find the equation of the circle.
(5) The area of a circle is I6n square units. If the centre of the circle is
(7,3), find the equation of the circle.
Find the equation of the circle whose centre is ( 4, 5) and
circumference is 8ti units.
2 2
Find the circumference and area of the circle x +y 6x8y+15 =
Find the equation of the circle which passes through (2, 3) and whose
centre is on xaxis and radius is 5 units.
Find the equation of the circle described on the line joining the points
(1,2) and (2, 4) as its diameter.
(10) Find the equation of the circle passing through the points (1, 0), (0,  1)
and (0, 1).
(11) Find the equation of the circle passing through the points (1, 1), (2, 1)
and (3, 2).
(12) Find the equation of the circle that passes through the points (4, 1) and
(6, 5) and has its centre on the line 4x + y= 16.
(13) Find the equation of the circle whose centre is on the line x = 2y and
which passes through the points ( 1, 2) and (3,  2).
(14) Find the cartesian equation of the circle whose parametric equations are
x = t cos9, y = t sin 9 and < 9 < 2n
2 2
(15) Find the parametric equation of the circle Ax +Ay =9
5.6. Tangent
5.6.1 Introduction
Let us consider a circle with centre at C and a straight line AB. This
straight line can be related to the circle in 3 different positions as shown in the
following figures.
(a)
(b)
(c)
Fig. 5.13
167
In figure (5.13 a), the straight line AB does not touch or intersect the circle.
In figure (5.13 b), the straight line AB intersects the circle in two points
and it is called a secant.
In figure (5.13 c), the straight line AB touches the circle at exactly one
point, and it is called a tangent. In otherwords, the limiting form of a secant is
called a tangent (Fig. 5.13d)
Definition : A tangent to a circle is a straight line which intersects
(touches) the circle in exactly one point.
5.6.2 Equation of the tangent to a circle at a point (x v y 1 )
Let the equation of the circle be
x 2 + y 2 + 2gx + 2fy + c = ...(1)
Let P(Xj, y j ) be a given point on it.
.. x 2 + y 2 + 2gx [ +2fy l +c = ...(2)
Let PT be the tangent at P.
The centre of the circle is C(— g, —J).
Fig. 5.14
Slope of the CP =
Since CP is perpendicular to PT, slope of PT =
Equation of the tangent PT is
yi+f
xi +g
x\+g
yi +/,
yy l = w(xXj)
x\ + 8
yyi
yi+f.
(X  Xj)
Cy  yO (yi+f) = (x *,) (x l + g)
(yyi)(y l +f) + (xx l )(x l + g) = o
=> yy l y 1 2 +fyfy l + [xx l x 2 + gxgx l ] = o
=> xx l + yy l +fy + gx = x 2 + y 2 + gx { +fy {
Add g*j +/y 1 + c on both sides
xx l +yy l + gx + gx 1 +fy+fy l + c = x l +y l +2gx 1 +2fy l + c
xx l + yy l + g(x + x±) + fly + y±) + c = is the required equation of the
tangent at (x^, y^)
168
Corollary:
2 2 2
The equation of the tangent at (x { , y^) to the circle x + y = a is
2
xx, + yy, = a .
2
Note: To get the equation of the tangent at (Xj, y^), replace x as xxp
2 x + x\ y + y\
y as yy^, x as — j — anc ^ y as — ? — * n tne ec l uat i ori °f tne circle.
5.6.3 Length of the tangent to the circle from a point (x v y 1 )
Let the equation of the circle be
x 2 + y 2 + 2gx + 2fy + c =
Let PT be the tangent to the circle from [ it— — ^ ' P<xj, y ( )
P(xj, jj) outside it. We know that the
coordinate of the centre C is ( g, —J) and
radius r = CT = '\jg 2 +f 2 c
From the right angled triangle PCT,
PT 2 = PC 2  CT 2
= ( Xl+ g) 2 + (y 1+f) 2 (g 2 + f 2 c)
= Xj +2gx l +g +y 1 +2fyj+f g f +c
= x 1 +yj +2gx 1 +2fy 1 + c
:. PT = a/xi + y\ + 2gx\ + 2fy\ + c , which is
the length of the tangent from the point (x^,y^)
2 2
to the circle x + y + 2gx + 2fy + c =
2
Note : (1) If the point P is on the circle then PT = (PT is zero).
2
(2) If the point P is outside the circle then PT > (PT is real)
2
(3) If the point P is inside the circle then PT < (PT is imaginary)
Corollary:
The constant c will be positive if the origin is outside the circle, zero if it is
on the circle and negative if it is inside the circle.
5.6.4 The condition for the line y = mx + c to be a tangent to the
circle x 2 + y 2 = a 2
2 2 2
Let the line y = mx + c be a tangent to the circle x +y =a at (x\, y{)
But the equation of the tangent at (x±, y^) to the circle
2 2 2 9
x + y = a is xxy + yy^ = a
169
Thus the equations y = mx + c and xxj + yy l = a are representing the same
straight line and hence their coefficients are proportional.
1 m c
>'l
x\
2
 a m
a
2 2 2
But (Xpjj) is a point on the circle x +y =a
2, 2 2
. . Xj +y, = a
2 2 2 2
am + a = c
4 2 4
a m a_
o + 1
= a
a 2 (m 2 + 1) = c 2
i.e.
2 2 2
c =a (1 + m ) is the required condition.
2 2 2
Note:(l)The point of contact of the tangent y = mx + c to the circle x +y = a is
 am a
J\jl +m 2 ' \jl +m 2 _
(2) The equation of any tangent to a circle is of the form
y = mx ± a Ajl + m
5.6.5 Two tangents can be drawn from a point to a circle
Let (x,, jj) be the given point. We know that y = mx±a\l + m is the
equation of any tangent. It passes through (xj, y,).
.". y, = mx, ±a\jl + m
y<  mx, = + a \jl + m
(y, mx,) =a (1 + m )
a 2 m 2 =
=> y.+mx, 2mx,y,  c
=> m (x,  a )  2mx,y~, + (y,  a ) =
This is a quadratic equation in 'm'. Thus 'm' has two values. But 'm' is the
slope of the tangent. Thus two tangents can be drawn from a point to a circle.
Note : (1) If (Xp yj) is an exterior point (lies outside) then both the tangents
are real and visible
170
Q(*> v 2 )
■>,)
Fig. 5.16
(2) If (jcj, yj) is an interior point (lies inside) the circle then both the
tangents are imaginary and hence not visible.
(3) If (xp yj) is a boundary point (lies on) then both the tangents
coincide and appears to be one.
5.6.6. Equation of the chord of contact of tangents from a point
to the circle
The general equation of the circle is
x 2 + y 2 + 2gx + 2fy + c = ...(1)
Let P(xj, >>j) be a point outside the circle.
Let the tangents from P(xj, y^) touch the
circle at Q(x 2 , y 2 ) and R(x 3 , y 3 )
The equation of the tangent PQ at Q (x 2 , y 2 ) * s
xx 2 + yy 2 + g(x + x 2 ) +/(y + y 2 ) + c =
The equation of the tangent PR at R(x 3 , y 3 ) is
xx 3 + xy 3 + g(x + x 3 ) +/(y + j 3 ) + c =
But (xj, yj) satisfy the equations (2) and (3)
•'• x \ x l + >'l>'2 + ^^l + ^ + ^>'l + J2) + c = ° and
XjX 3 + y x y 3 + g(x x + x 3 ) +/ijj + y 3 ) + c =
But equations (4) and (5) show that (x 2 , y 2 ) and (x 3 , y 3 ) lie on the line
xxj + yy 1 + g(x + Xj) +/(j + y t ) + c =
Hence the straight line xxj + yy^ + g(x + Xj) +f(y + y^) + c = represents
the equation of QR, chord of contact of tangents from (x J: y^).
Example 5.46: Find the length of the tangent from (2, 3) to the circle
x 2 + /4x3;y+12 = 0.
Solution:
2 2
The length of the tangent to the circle x +y + 2gx + 2fy + c = from the
(2)
(3)
(4)
(5)
point (x v yj) is y^ 2 + y 2 + 2gx x + 2fy x + c
:. Length of the tangent to the given circle is ~\lx\
= V2 2 + 3
2 + yi 2 4xi3yi + 12
4.23.3+12
= ^4 + 989+ 12
= ^/8 = 2^/2 units
171
Example 5.47: Show that the point (2, 3) lies inside the circle
x 2 + y 2 6x8y + 12 = 0.
Solution:
The length of the tangent PT from P(xj, jj) to the circle
x + y + 2gx +2fy + c = is
PT= \jx l 2 + y l 2 + 2gx l + 2fy l + c
PT 2 = 2 2 + 3 2  6.2  8.3 + 12 = 4 + 9  12  24 + 12
=  11 <
The point (2, 3) lies inside the circle
2 2
Example 5.48: Find the equation of the tangent to the circle x +y =25 at (4, 3).
Solution:
2 2
The equation of the circle is x +y = 25 .
The equation of the tangent at (xj, jj) is xxj + yy l = 25. Here (xj, y^ = (4, 3).
.". The equation of the tangent at (4, 3) is 4x+3y = 25
2 2
Example 5.49: If >>=3x+c is a tangent to the circle x +y =9, find the value of c.
Solution:
The condition for the line y = mx + c to be a tangent to
2 2 2 / 2
x + y = a isc=±a'\jl+m
Here a = 3, m = 3
.. c = ± 3 VTO
Example 5.50: Find the equation of the tangent to
x 2 + y 2  4x + Ay  8 = at ( 2,  2)
Solution:
The equation of the tangent at (x\, y\) to the given circle is
fx + x{\ fy + y{\
xx 1+}7l 4^J +4(2J 8 =
xxj + yy^  2 (x + Xj) + 2(y + y^)  8 =
At ( 2,  2), the equation of the tangent is
 2x  2y  2 (x  2) + 2(y  2)  8 =
=>  4x  8 =
=> x + 2 = is the required equation of the tangent.
Example 5.51: Find the length of the chord intercepted by the circle
2 2
x +y 2xy + I =0 and the line x  2y = 1 .
172
Solution:
To find the end points of the chord, solve the equations of the circle and
the line. Substitute x = 2y + 1 in the equation of the circle.
(2y + l) 2 + y 2  2(2y + 1)  y + 1 =
Ay 2 + Ay + 1 + y 2  Ay  2  y + 1 =0
5y 2 y =
y =
x= 1
.•. The two end points are (1, 0) and [T ,T
7 1
y(5y  l) =
l
? = 5
7
x ="
.". Length of the chord = A / 1 1
s) H°i) 2
T\ 2
25 + 25
: V5
units
Example 5.52: Find the value of/? if the line 3x + Ay  p = is a tangent to the
2 2
circle x +>> =16.
Solution:
2 2 2
The condition for the tangency isc = a (1 + m ) .
2 3d
Here a =16, m = — t, c=^
c 2 = a 2 (l+m 2 )
l^= 16 l 1+ ^J =25
// = 16 x 25
:.p = ± 20
Example 5.53: Find the equation of the circle which has its centre at (2, 3) and
touches the xaxis.
Solution:
Let P be a point on xaxis where it touches the circle.
Given that the centre C is (2, 3) and P is (2, 0)
Y
•=CP = V(22) 2 + (30) 2 =3
2 2 2
The equation of the circle is (x  h) +(yk) =r
(x2) 2 + (y3) 2 = 3 2
o
P(2. 0)
X
Fig. 5.17
x 2 + y 2  Ax  6y + A =
173
(1
(2
(3
(4
(5
(6
(7
(8
(9
(10
(11
(12
(13
(14
EXERCISE 5.6
Find the length of the tangent from (1, 2) to the circle
x + y 2  2x + Ay + 9 =
x 2 + y 2 + 2x  A = and
qual.
2 2
Find the equation of the tangent to the circle x +y 4x + 8y  5 = at
2 2
Is the point (7,  11) lie inside or outside the circle x +y  10x = 0?
the po
2 2
or inside the circle x +y  5x + 2y  5 =
Prove that the tangents from (0, 5) to the circles
x 2 + y 2 + 2x  A = and
2 2
x +y  )> + 1 = are equal.
Find th
(2, 1).
Is the f
Determine whether the points ( 2, 1), (0, 0) and (4,  3) lie outside, on
2 2
or inside the circle x +y  5x + 2y  5 =
Find the coordinates of the point of intersection of the line x + y = 2
2 2
with the circle x + y = 4
2 2
Find the equation of the tangent lines to the circle x + y = 9 which are
parallel to 2x + y  3 =
Find the length of the chord intercepted by the circle
x 2 + y 2  14jc + Ay + 28 = and the line x  ly + 4 =
Find the equation of the circle which has its centre at (5, 6) and touches
(i) xaxis (ii) jaxis
2 2
Find the equation of the tangent to x +y  2x  lOy + 1 = at ( 3, 2)
2 2
Find the equation of the tangent to the circle x +y =16 which are
(i) perpendicular and (ii) parallel to the line x + y = 8
2 2
Find the equation of the tangent to the circle x +y  4x + 2y  21 =
at (1,4).
Find the value of/? so that the line 3x + Ay p = is a tangent to
x 2 + y 2  64 =
Find the coordinates of the middle point of the chord which the circle
2 2
x +y +2x + )'3 = 0cuts off by the line y = x  1 .
5.7. Family of circles
Concentric circles:
Two or more circles having the same centre are called concentric circles.
Circles touching each other:
Two circles may touch each other either internally or externally. Let C 1?
C 2 be the centers of the circle and r^, r 2 be their radii and P the point of
contact.
174
Case (1):
The two circles touch externally.
The distance between their centres
is equal to the sum of their radii.
(i.e.) CjC 2 = r j + r 2
Case (2) The two circles touch
internally:
The distance between their
centres is equal to the difference
of their radii.
Cj C 2  CjP
Fig. 5.18
Fig. 5.19
Orthogonal circles:
Definition: Two circles are said to be orthogonal if the tangent at their
point of intersection are at right angles.
Condition for two circles cut orthogonally
Let the two circles be
2 2
x + y + 2gyX + 2/jj + Cj = and
2 2
x + y + 2g 2 x + 2f 2 y + c 2 = and cut each
other orthogonally.
Fig. 5.20
Let A and B be the centres of the two circles
•'• A is ( g v  /j) and B is ( g 2 ,  f 2 ) r x = ^g { 2 +f { 2  c\ and
r 2 = ^82 2 +f2 2 C2
In the right angled triangle APB, AB 2 = AP 2 + PB 2
i.e.
(g l +g 2 ) 2 + (f 1 +f 2 ) 2 = g l 2 + fl 2 c 1 + g 2 2 +f 2 2 c 2
=> g\ +8 2  2 8i8 2 + fi + h  2 /l/2 = 8[ +/i ~ c i + g2 + fl ~ c 2
=> 2g i g 2 2fJ 2 = c l c 2
i.e. 2g l g 2 + 2f l f 2 = c l + c 2
is the required condition for orthogonality.
2 2
Example 5.54: Show that the circles x +y 4x + 6y + 8 = and
2 2
x + y  10x  6y + 14 = touch each other.
175
2 2
x +y 4x 6y 9 = and passing through the point ( 4,  5).
Solution:
The given circles are
x 2 + y 2 4x + 6y + 8 = ... (1)
and x 2 + y 2  lOx  6y + 14 = ...(2)
(1) => gl =  2 /j = 3, Cj = 8. Centre is A(2,  3)
radius ^ = ^g\ 2 +f\ 2 ci = yJ4 + 9  8 = a/5
(2) => g 2 =  5,/ 2 =  3, c 2 = 14. Centre is B(5, 3)
radius r 2 = ^25 + 914 = a/20 = 2^/5
Distance between A and B = "\/(2  5) 2 + ( 3  3) 2
= V9 + 36 = ^45 =3a/5
= r l + r 2
:. The circles touch each other.
Example 5.55: Find the equation of the circle, which is concentric with the
circle
x
Solution:
2 2
The given circle isx +y 4x6y 9 =
Centre ( g, f) is (2, 3)
The circle passes through the point ( 4,  5).
.. radius = "\/(2 + 4) 2 + (3 + 5) 2 = a/36 + 64 = ^100 = 10
2 2 2
The equation of the circle is (xh) +(yk) =r
Here (h, k) = (2, 3), r = 10
.. (jc  2) 2 + (y  3) 2 = 10 2
2 2
x + y  4x  6y  87 = is the required equation of the circle.
2 2
Example 5.56: Prove that the circles x +y  8x + 6y  23 = and
2 2
x +>>  2x  5y + 16 = are orthogonal.
Solution:
The equations of the circle are
x 2 + y 2 8x + 6y23 = ...(1)
x 2 + y 2 2x5y+l6 = ...(2)
176
(1) => gl = 4, / 1 = 3, c 1 = 23
5
(2) => g 2 =  1, / 2 =  2 , c 2 = 16
Condition for orthogonality is 2g 1( g 2 + 2/j/ 2 =Cj + c 2
2g { g 2 + 2f x / 2 = 2(4) ( 1) + 2 (3) (§)  8  15 =  7
C j + c 2 = 23 + 16 = 7
2^2 + 2 /l/2 = c i +c 2
.". The two circles cut orthogonally and hence they are orthogonal circles.
Example 5.57:
Find the equation of the circle which passes through the point (1, 2) and
2 2 2 2
cuts orthogonally each of the circles x + y = 9 and x + y 2x + 8y 1 =
Solution:
2 2
Let the required equation of the circle be x +y + 2gx + 2fy + c = •■•(!)
The point (1,2) lies on the circle
1 + 4 + 2g + 4/+ c =
2g + 4f+c = 5 ...(2)
2 2
The circle (1) cuts the circle x +j = 9 orthogonally.
2«i« 2 + 2 /l/2 = c \ + c 2
=>
2g(0)+2f(0) = c9
:. c = 9
...(3)
Again the circle (1)
cutsx
2
+ y 2x + 8y  7 = orthogonally.
.. 2g(l) + 2/(4) = C 7
=>
2g + 8f =97 = 2
=>
S + 4/=l
...(4)
(2) becomes
2g + 4/=14
••• g + 2/=7
...(5)
(4) + (5) =>
6/= 6 =>/= 1
(5) =>
g2=7 => g = 5
2 2
.". The required equation of the circle is x +y 10x2y + 9 =
177
EXERCISE 5.7
(1) Show that the circles x + y 2x + 6y + 6 =
2 1
and* +y  5x + by + 15 = touch each other.
(2) Show that each of the circles x +y + 4y  1 = 0, x +y +6x + y + 8=0
2 2
and x +y  Ax  Ay  37 = touches the other two.
(3) Find the equation of the circle concentric with the circle
2 2
x +y  2x  6y + A = and having radius 7.
(4) Find the equation of the circle which is concentric with the circle
2 2
x + y  8x + 12y + 15 = and passes through the point (5, 4)
2 2
(5) Show that the circle x + y  8x  6;y + 21 = is orthogonal to the
circle x 2 + y 2  2y  15 =
(6) Find the circles which cuts orthogonally each of the following circles
(i) x 2 + y 2 + 2x + Ay + 1 = 0, x 2 +y 2  Ax+3 = and x 2 + y 2 + 6y + 5 =
(ii) x 2 + y 2 + 2x + I7y + A = 0, x 2 + y 2 + Ix + 6y + 1 1 =
and x 2 + y 2 x + 22y + 3 =
(7) Find the equation of the circle which passes through (1,  1) and cuts
2 2
orthogonally each of the circles x +y + 5x5y + 9 = and
x 2 + y 2  2x + 3y  1 =
(8) Find the equation of the circle which passes through (1, 1) and cuts
2 2
orthogonally each of the circles x +y &x2y+ 16 =
and x 2 + y 2  Ax  Ay  1 =
178
6. TRIGONOMETRY
6.1 Introduction:
Trigonometry is one of the oldest branch of Mathematics. The word
trigonometry means "triangle measurement". In olden days trigonometry was
mainly used as a tool for use in astronomy. The early Babylonians divided the
circle into 360 equal parts, giving us degrees, perhaps because they thought that
there were 360 days in a year.
The sine function was invented in India, perhaps around 300 to 400 A.D.
By the end of ninth century, all six trigometric functions and identities relating
them were known to the Arabs.
In its earlier stages trigonometry was mainly concerned with establishing
relations between the sides and angles of a triangle, but now it finds its
application in various branches of science such as surveying, engineering,
navigation etc. For every branch of higher Mathematics a knowledge of
trigonometry is essential.
6.1.1 Angles:
An angle is defined as the amount of
rotation of a revolving line from the initial
& \ positive angle
position to the terminal position. \r ft ( anti clockwise)
Counterclockwise rotations will be
called positive and the clockwise will be /— negative angle
called negative. / (clockwise)
Consider a rotating ray OA with its a* y'
end point at the origin O. p[„ <5 1
The rotating ray OA is often called the terminal side of the angle and the
positive half of the xaxis (OX) is called the initial side.
The positive angle 9 is XOA (counterclockwise rotation)
The negative angle 9 is XOA' (clockwise rotation)
Note : 1 . one complete rotation (counter clockwise) = 360° = 360 degree
2. If there is no rotation the measure of the angle is 0°.
6.1.2 Measurement of angles:
( i V h
If a rotation from the initial position to the terminal position is I ttt: I of
the revolution, the angle is said to have a measure of one degree and written as
1°. A degree is divided into minutes, and minute is divided into seconds.
179
i.e. 1 degree (1°) = 60 minutes (60')
1 minute (1') = 60 seconds (60")
In theoretical work another system of measurement of angles is used which
is known as circular measure. A radian is taken as the unit of measurement.
6.1.3 Radian measure:
Definition:
One radian, written as l 6 is the measure of an angle
subtended at the centre O of a circle of radius r by an arc
of length r.
Note
L. To express the measure of an angle as a real
number, we use radian measure.
2. The word "radians" is optional and often omitted. Thus
is given for a rotation, it is understood to be in radians.
3. V in l c indicates the circular measure.
6.1.4 Relation between Degrees and Radians
Since a circle of radius r has a circumference of 2nr, a
circle of radius 1 unit (which is referred to as an unit circle)
has circumference 2it. When 9 is a complete rotation, P
travels the circumference of an unit circle completely.
Fig. 6. 3
If 9 is a complete rotation (counterclockwise) then 9 = 2it radian. On the
other hand we already know that one complete rotation (counterclockwise) is
360°, consequently, 360° = 2 it radians or 180° = n radian. It follows that
,„ n , 180 °
1° = yjyT radian and ■
71
= 1 radian. Therefore 1° = 0.01746 radian (app.) and
7
1 radian = 180° x^ =57
16' (app.).
Conversion for some special angles:
degrees
30°
45°
60°
90°
180°
270°
360°
Radians
6
%
4
71
3
n
2
%
3ic
2
2ic
(Table 6.1)
Example 6.1: Convert
3ic 1
(i) 150° into radians (ii) 7 into degrees (iii) t radians into degrees.
180
Solution:
(i)
150° =
n 5
150 x Toq radians = t n
(ii)
3ji
7 radians =
3 ; x 18 °° = 135°
4 n
(iii)
t radians =
1 18 ° l i«n 7
4 X V = 4 x 180x 2l
6.1.5 Quadrants
Let X'OX and YOY' be two lines at right angles
to each other as in the fig. (6.4) we call X'OX and
YOY' as xaxis and yaxis respectively.
Clearly these axes divide the entire plane into four
equal parts, called quadrants.
14° 19' 5"
IV
*x
"Y
Fig. 6. 4
The parts XOY, YOX', X'OY' and Y'OX are known as the first, the
second, the third and the fourth quadrant respectively.
Angle in standard position:
If the vertex of an angle is at O and its initial side lies along xaxis, then the
angle is said to be in standard position.
Angle in a Quadrant:
An angle is said to be in a particular quadrant, if the terminal side of the
angle in standard position lies in that quadrant.
Example 6.2: Find the quadrants in which the terminal sides of the following
angles lie.
(ii)  300°
©■
x«
60°
Y
«)
*\
Fig. 6. 5 a
From Fig (6.5a)
the terminal side of
 60° lies in IV
quadrant.
Y
Fig. 6.5 b
From Fig (6.5b) the
terminal side of  300°
lies in I quadrant.
v
Fig. 6.5 c
From Fig (6.5c)
1295° = 3x360°+180° + 35°
The terminal side lies in III
quadrant.
181
EXERCISE 6.1
(1) Convert the following degree measure into radian measure.
(i) 30° (ii) 100° (iii) 200° (iv)  320°
(v)  85° (vi) 7° 30'
(2) Find the degree measure corresponding to the following radian measure
(i,(f) (ii)^) (iii) 3 (iv) @f)
(3) Determine the quadrants in which the following degrees lie.
(i)380° (ii) 140° (iii) 1100°
6.2. Trigonometrical ratios and Identities
6.2.1 Trigonometrical ratios:
In the coordinate plane, consider a point A
on the positive side of xaxis. Let this point
revolve about the origin in the anti clockwise
direction through an angle 9 and reach the point P.
Now XOP = 9. Let the point P be (x, y). Draw
PL perpendicular to OX.
The triangle OLP is a right angled triangle, in
which 9 is in standard position. Also, from the
AOLP, we have
OL = x = Adjacent side ; PL = y = opposite side ;
OP = ^jx 2 + y 2 = Hypotenuse (= r > 0)
The trigonometrical ratios (circular functions) are defined as follows :
y
The sine of the angle 9 is defined as the ratio ~. it is denoted by sin9.
o
e
3 N
L A
j» 5i
Fig. 6. 6
i.e.
and
sin 9 =
cos 9 =
tan 9 =
1
r
x
r
1
x
cosecant value at 9 = ~~ = cosec 9 ; y ^
r
secant value at 9 = ~ = sec 9 ; x *■
cotangent value at 9 = ~ = cot 9 ; y ^
Note : 1. From the definition, observe that tan 9 and sec 9 are not defined
if x = 0, while cot 9 and cosec 9 are not defined if y = 0.
2. cosec 9, sec 9 and cot9 are the reciprocals of sin9, cos 9 and tan9
respectively.
182
Example 6.3: If (2, 3) is a point on the terminal side of 9, find all the six
trigonometrical ratios.
Solution:
P(x, y) is (2, 3) and it lies in the first quadrant.
.". x = 2, y = 3
■ = ^x 2 + y 2 =^A + 9 = >/l3
x 2
C0Se = r=VT3
cosec 9 =
13
sec 9
Vl3
v 3
tan9=^=2
„ 2
cot 9 = t
PR3)
3
\
Fjg. 6. 7
Note : 1. From example (6.3), we see that all the trigonometrical ratios are
positive when the terminal side of angle 9 lies in first quadrant.
Now, let us observe the sign of trigonometrical ratios if the point on the
terminal side of 9 lies in the other quadrants, (other than the first quadrant).
Example 6.4: If ( 2,  3) is a point on the terminal side of 9. Find all the six
trigonometrical ratios.
Solution:
P(x, y) is ( 2,  3) and it lies in the third o
quadrant /
••• x = 2 , y =  3 ; ( _2, 3 )
r = \jx 2 + y 2 =A /4T9 =y/l3 Fig. 6. 8
TV
sin 9 = T r =
.1.
= = ve;cose =  r = ^= = ve;tmQ=^ = 
yi3
3 '
■_3
2
= 2 = + ve
cosec 9 = ~ = * 1 r~ =  ve ; sec 9 = ~ =
r
x
13 A 2
— =ve ; cot 9 = ~r
= 3 = + ve
As example illustrates, trigonometric functions may be negative. For
y r
instance, since r is always positive, sin 9 = and cosec 9 = ~ have the same
sign as y. Thus sin 9 and cosec 9 are positive when 9 is in the first or second
quadrants, and negative when 9 is in the third or fourth quadrants. The signs of
the other trigonometric functions can be found similarly. The following table
indicates the signs depending on where 9 lies.
183
Quadrants
Functions
I
II
Ill
IV
Sin
+
+


Cos
+


+
Tan
+

+

Cosec
+
+


Sec
+


+
Cot
+

+

II
sin
cosec
I
All
III
tan
cot
IV
cos
sec
Table (6.2)
6.2.2 Trigonometrical ratios of particular angles:
Let X'OX and YOY' be the coordinate axes.
With O as centre and unit radius draw a circle
cutting the coordinate axes at A, B, A' and B' as
shown in the figure.
Suppose that a moving point starts from A and
move along the circumference of the circle. Let it
cover an arc length. 9 and take the final position P.
Let the coordinates of this point be P(x, y). Then
by definition, x = cos9 and y = sin9.
We consider the arc length 9 to be positive or negative according as the
variable point moves in the anti clockwise or clockwise direction respectively.
Range of cos 9 and sin9 :
Since for every point (x, y) on the unit circle, we have
 1 < x < 1 and  1 < y < 1, therefore  1 < cos 9 < 1 and  1 < sin9 < 1
n 3n
Values of cos9 and sin9 for 9 = 0, ~ » n, ~y and 2ti.
We know that the circumference of a circle of unit radius is 2rc. If the
moving point starts from A and moves in the anti clockwise direction then at the
points A, B, A', B' and A the arc lengths covered are 9 = 0, y , n, ~j~ and 2%
respectively.
Also, the coordinates of these points are:A(l, 0), B(0, 1), A'(— 1, 0),
B'(0,l)andA(l,0)
At the point:
A(l, 0), 9 = => cosO =1 and sin =
184
B (0, 1),
= f
=>
K
COST
=
and
sin T =1
A' (1, 0),
Q = n
=>
cos n
= 1
and
sin n =
B' (0,  1),
93f
=>
cos 3 y
=
and
sin 3 x =
A (1,0),
Q = 2n
=>
cos 2tc
= 1
and
sin 2n =
6.2.3 Trigonometrical ratios of 30°, 45° and 60°
Consider an isosceles rightangled triangle
whose equal sides are 1. Its hypotenuse is
^/l 2 + l 2 = ^2 .Its base angle is 45°.
.. sin 45° = ;= ; cos 45° = 7= ; tan45° = 1
cosec 45° = ^2 ; sec 45° = ^2 ; cot 45° = 1 Fjg. 6. 10
Opposite side = 1
adjacent side = 1
hypotenuse = yJ2
Consider an equilateral triangle ABC of side 2 units. Each of its angle is
60°. Let CD be the bisector of angle C. Then angle ACD is 30°. Also AD = 1
ndCD = V2 2 l 2 =
V3
. Now in the right ar
igled triangle ACD
For 30°
For 60°
opposite side = 1
opposite side = y]3
adjacent side = \J3
C
adjacent side = 1
hypotenuse = 2
hypotenuse = 2
sin 30° = 2
2 /3(f
A
sin 60° = \
™n "\/3
£<&
,
cos 30° = 2
A 1 D 1 B
cos 60° = 2
tan 30°=^
Fig. 6.11
tan 60° = a/3
.'. cosec 30° = 2
.". cosec 60° = —fc
••• sec 30 ° = J3
:. cot 30° = ^3
sec 60° = 2
, cot 60° = ^
185
e
K
n
7t
71
n
371
2tc
6
4
3
2
2
sin9
1
2
1
V2
2
1
1
cos 9
1
2
1
1
2
1
1
tan 9
1
V3
1
V3
CO
— 00
Table 6.3
Important results:
For all values of 9, cos ( 9) = cos 9 and sin ( 9) =  sin 9
Proof:
Let X'OX and YOY' be the coordinate axes. With O as centre and unit
radius draw a circle meeting OX at A. Now let a moving point start from A and
move in anti clockwise direction and take the final position P(x, y) so that arc
AP = 9.
On the other hand, if the point starts from A and moves in the clockwise
direction through the arc length AP' equal to arc length AP. Then arc AP'=  9.
Pfejj)
Fig. 6. 12
Thus  AOP = 9 and  AOP' =9
From the coordinate geometry, we know that the coordinates of P' are
(x,y).
Clearly, cos9 and cos(9) are respectively the distances of points P and
P' from y axis and clearly each one of them is equal to x.
:. cos ( 9) = cos 9
Clearly, sin 9 and sin( 9) are respectively the distances of points P and
P' from xaxis. As sin9 = y and sin (9) = — y, we have sin(9) =  sin 9
Deductions cosec (9) =  cosec 9 ; sec (9) = sec 9
tan (9) =  tan 9 ; cot (9) =  cot 9
186
6.2.4 Tratios of (90° ± 9), (180° ± 9), (270° ± 9) and (360° ± 9):
It is evident that, when 9 is a small angle (0 < 9 < 90°), then
90°  9, 90° + 9 etc., are in the quadrants as given below:
Angle
Quadrant
90° 0
Ql (first quadrant)
90° + 9
Q2
180° 
Q 2
180° +
Q 3
270° 
Q3
270° + 9
Q 4
360°  9 ; also equal to " 9"
Q 4
360° +
Qi
Table 6.4:
Let P(a, (3) be a point in the first quadrant. Let XOP = 9
/. sin 9 = 7777 ; cos9 = 7777 ; tan 9 =
OP a
OP
n 0P n 0P ^ a
cosec 9 = ~tt ; sec 9 = — ; cot9 = 77
(3 a (3
Tratios of (90° 9)
Let Q be a point in the first quadrant such
that
Fig. 6. 13
XOQ = 90°  and OQ = OP.
Let PA and QB be perpendicular to OX and OY respectively.
Then AOAP = AOBQ and Q is (p\ a). Hence
sin (90°  0) =
cos (90°  9) =
y coordinate of Q a
OQ
OP
x coordinate of Q (3
OQ
OP
= cos 9
= sin 9
, nnn ., y coordinate of Q a
tan < 90 V=x coordinate of Q = p = cot9
Similarly, cosec (90°  9) = sec 9
sec (90°  0) = cosec 9
cot (90°  0) = tan 9
187
Tratiosof(90° + 9)
Let R be a point in the second quadrant such
that XOR =90° + 9andOR = OP
Let RC be perpendicular to x axis.
Then AOAP = ARCO and R is ( (3, a), Hence
y coordinate of R
sin (90° + 9) = l q^ :
Y
R(M
e\
P
C
O a t
v x
cos (90° + 9) =
x coordinate of R
OR
y coordinate of R
tan ( 90 ° + 9 )  .coordinate of R
Similarly, cosec (90° + 9) = sec 9
sec (90° + 9) =  cosec 9
cot (90° + 9) =  tan 9
Tratiosof(180°9)
Let S be a point in the second quadrant such that
XOS =180° 9 and OS =OP
Draw SD perpendicular to xaxis
Thus A OAP = A ODS and S is ( a, (3). Hence
Fig. 6. 14
a a
Op" = cos9
: ^p =  sin 9
a
: ^jt =  cot 9
Y
S(
«P)
A
P(«,P)
V
Xe
D O
A X
sin (180°
cos (180°
tan (180°
y coordinate of S
9) = QS
9) =
x coordinate of S
OS
y coordinate of S
' ~ x coordinate of S
OP
 a
OP
A.
 a
Fig. 6. 15
= sin 9
=  cos 9
= tan 9
Similarly, cosec (180°  9) = cosec 9
sec (180° 9) = sec 9
cot (180° 9) = cot 9
Tratiosof(180° + 9)
Let T be a point in the third quadrant such that
XOT = 180° + 9 and OT = OP
Draw TE perpendicular to xaxis
Then AOAP = AOET and T is ( a,  (3).
Hence
P(aP;
188
sin (180° + 9) =
cos (180° + 9) =
tan (180° + 9) =
y coordinate of T (3
OT = OP
x coordinate of T  a
OT = OP
y coordinate of T (3
=  sin 9
=  cos 9
= tan 9
Similarly, cosec (180° + 9) =
sec (180° + 9) =
x coordinate of T  a
 cosec 9
 sec 9
cot (180° + 9) = cot 9
Remark: To determine the trigonometric ratios of any angle, follow the
procedure given below
n
(i) Write the angle in the form & x + 9 ; k e Z.
(ii) Determine the quadrant in which the terminal side of the angle lies,
(iii) Detrmine the sign of the given trigonometric function in that
, • S A ,
particular quadrant, using ™ ^ rule.
(iv) If k is even, trigonometric function of allied angle equals the same
function of 9.
(v) If k is odd, then adopt the following changes:
sine <> cos ; tan <>■ cot ; sec <> cosec
Trigonometrical ratios for related angles
Xngle
function"
e
90 e
90 + 9
1809
180+9
2706
270+9
3609
or
9
sin
sin9
cos 9
cos 9
sin 6
 sin 9
cos e
cos 9
sin6
cos
cos 8
sin 9
sin 9
cos 9
cos 9
sine
sin 9
cos 9
tan
tanG
cot 9
cote
tan 9
tan 9
cote
cot 9
tan9
cosec
cosecG
sec 9
sec 9
cosec 9
cosec9
sec 6
sec 9
cosec9
sec
sec 9
cosec 9
cosec9
sec 9
sec 9
cosece
cosec 9
sec 9
cot
cote
tan 9
tan9
cote
cote
tan6
tan 9
cot 9
Table 6.5
Note : Since 360° corresponds to one full revolution, sine of the angles
360 o +45 o ;720 o +45 o ;1080 o +45 o are equal to sine of 45°. This is so for the other
trigonemetrical ratios. That is, when an angle exceeds 360°, it can be reduced to
an angle between 0° and 360° by wiping out integral multiples of 360°.
189
Example 6.5:
Simplify : (i) tan 735° (ii) cos 980° (iii) sin 2460° (iv) cos (870°)
(v) sin (780°) (vi) cot (855°) (vii) cosec 2040°(viii)sec ( 1305°)
Solution:
(i) tan (735°) = tan (2 x 360° + 15°) = tan 15°
(ii) cos 980° = cos (2 x 360° + 260°) = cos 260°
= cos (270°  10°) =  sin 10°
(iii) sin (2460°) = sin (6 x 360° + 300°) = sin (300°)
= sin (360°  60°)
=  sin 60°
_ 2
(iv) cos ( 870°) = cos (870°) = cos (2 x 360° + 150°)
= cos 150 = cos (180°  30°)
= cos 30° = \
(v) sin ( 780°) =  sin 780°
=  sin (2 x 360° + 60°)
=  sin 60° =  *2
(vi) cot ( 855°) =  cot (855°) =  cot (2 x 360° + 135°)
=  cot (135°) =  cot (180°  45°)
= cot 45° = 1
(vii) cosec (2040°) = cosec (5 x 360° + 240°) = cosec (240°)
= cosec (180° + 60°) =  cosec (60°)
= "^
(viii) sec ( 1305°) = sec (1305°) = sec (3 x 360° + 225°)
= sec (225°) = sec (270°  45°)
=  cosec 45° =  ^2
,,,„.,.* cot (90° 9) sin (180° + 9) sec (360°  9)
Example 6.6: Simplify : — —  e)   e) cqs (9QO  Q)
190
. tan 9 ( sin 9) (sec 9)
Solution: The given expression = — zrrr — : — rr
tan 9 (sec 9) ( sin 9)
= 1
Example 6. 7:
Without using the tables, prove that sin 789° sin 489° + cos 129° cos69° = ~
Solution: sin 780° = sin (2 x 369° + 69°) = sin 69° = ^
sin 489° = sin (369° + 129°)
= sin 129° = sin (189°  69°) = sin 69° = \
cos 129° = cos (189°  69°) =  cos 69° =  j \ cos69° = ^
L.H.S.  2 2 2 2
\/3 \/3 I I
2 • 2
= 4 "4 =2 RHS 
6.2.5 Special properties of Trigonometrical functions:
Periodic function:
A function fix) is said to be a periodic function with period a if
fix + a)=fix). The least positive value of a is called the fundamental period of
the function.
All the circular functions (trigonometrical functions) are periodic
functions.
For example,
sin (x + 2n) = sin x ; sin (x + 4%) = sin x ; sin (x + 6n) = sin x
sin (x + 2nn) = sin x, n e Z
Here a =  6n,  4ti,  2n, 0, 2n, 4n, ... .But the fundamental period
must be the least positive quantity. Therefore a = 2n is the fundamental period.
Thus sine function is a periodic function with fundamental period 2n.
Similarly one can prove that the functions cos x, cosec x and sec x are also
periodic functions with fundamental period 2n while tan x and cot x are
periodic with fundamental period n.
191
6.2.6 Odd and even functions:
We know that, if fix) =fi x), then the function is an even function and if
fi x) = fix) then the function is an odd function.
Consider fix) = sin x ; fi x) = sin (%) =  sin x = fix) i.e. fix) = fi x)
.". sin x is an odd function. Similarly we can prove that cosec x, tan x and cot x
are odd functions.
Consider fix) = cos x ;/( x) = cos ( x) = cos x =fix). .'. cosx is an even
function. Similarly we can prove sec x is an even function.
Note : We can read more about odd and even function in Chapter 7.
EXERCISE 6.2
(1) If sin 9 = y2 , find the value of
sec (360°  9) . tan (180°  9) + cot (90° + 9) sin (270° + 9)
(2) Express the following as functions of positive acute angles: 
(i) sin ( 840°) (ii) cos (1220°) (iii) cot ( 640°) (iv) tan (300°)
(v) cosec (420°) (vi) sin ( 1110°) (vii) cos ( 1050°)
,„, „ , sin 300° . tan 330° . sec 420° fl
(3) Prove that — ^ . cos 210 ° . cosec 315° = ~ V3
71
(4) Prove that i 1+cot a  sec I a + ~ I ( 1 1 + cot a + sec I a + ~ I ( =2 cot a
(5) Express the following as functions of A :
(i)sec[A^"J (ii) cosec [A ^ J (iii) tan I A  y
(iv) cos (720° + A) (v) tan (A + ti)
,, N „ , sin (180°+A) . cos (90°A) . tan (270°A) . A lk
(6) Prove that , C/irio — tt ,,,.. .. ,,„ no .. = sin A cos A
sec (540° A) cos (360°+A) cosec (270°+A)
(7) Prove that sin 9 . cos 9 i sin I y  9 J . cosec 9 + cos I y  9 J sec 9 r = 1
(8) Find the values of :
(i) cos (135°) (ii) sin (240°) (iii) sec (225°) (iv) cos ( 150°)
5tc ( 5ti
(v) cot (315°) (vi) cosec ( 300°) (vii) cot j (viii) tan
(9) If A, B, C, D are angles of a cyclic quadrilateral prove that
cosA + cosB + cos C + cos D = 0.
192
(10) Find the values of the following expressions :
(i) tan 2 30° + tan 2 45° + tan 2 60° (ii) sin t . cos j + cos t . sin t
II Jl Jl Jl
(iii) cost, cost sinT.sinT (iv) cos 45°.cos60°sin 45°. sin
60°
(v) tan 2 60° + 2 tan 2 45°
(vi) tan 2 45° + 4 cos 2 60°
(vii) cot 60° . tan30° + sec z 45° . sin 90°
(viii) tan 2 60° + 4 cot 2 45° + 3 sec 2 30° + cos 2 90°
(ix) tan 2 30° + 2 sin 60° + tan 45°  tan 60° + cos 2 30°
(x) 2 sin 2 60°  j sec 60° tan 2 30° + 5 sin 2 45° . tan 2 60°
(11) If cos 9 =  o and tan 9 > show that —f= ~ —  — : — r = 3.
z y3 cos 9  3 sin 9
6.2.7 Trigonometrical identities:
2 2
As in variables, sin 9 . sin 9 = (sin 9) . This will be written as sin 9.
Similarly
2 ^
tan 9 tan 9 = tan 9 etc. We can derive some fundamental trigonometric
identities as follows:
Consider the unit circle with centre at the origin O. Let
P(x, y) be any point on the circle with XOP = 9.
Draw PL perpendicular to OX. Now, triangle OLP is a
right angled triangle in which (hypotenuse) OP = r = 1
unit, and x and y are adjacent and opposite sides
respectively.
x y y
Now we have cos 9 = j = x and sin 9 = 7" = y; tan 9 = ~_
and r 2 = x 2 + y 2 = 1
2 2 2
From AOLP, we have x +y = r =1
Fig. 6.17
i.e.
x 2 + y 2 = cos 2 9 + sin 2 9 = 1
1 + tan z 9
2._ 1+ £.£t£.(iy_(_L_p.._j
x x
..2 .2 , ..2
Vcos 9
sec z 9
1 + COt2e = 1+X ~2 JL f L = g) 2 = {^tof = sec 2 9
Thus we have the identities
193
sin 2 9 + cos 2 9 = 1
1 + tan 2 9 = sec 2 9
1 + cot 2 9 = cosec 2 9
From these we also have
sec 2 9  tan 2 = 1
2 2
cosec 9  cot 9=1
Example 6.8: Show that cos A  sin A = 1  2 sin A
Solution:
cos 4 A  sin 4 A = (cos 2 A + sin 2 A) (cos 2 A  sin 2 A)
= cos 2 A  sin 2 A = 1  sin 2 A  sin 2 A
= 1  2sin z A
2 2 2 2
Example 6.9: Prove that sec A + cosec A = sec A . cosec A
Solution:
2 2 '
sec A + cosec A =
2 2
cos A sin A
sin A + cos A
~~ 2* ■ 2* _ 2* • 2*
cos A . sin A cos A . sin A
2 2
= sec A . cosec A
Example 6.10: Show that cosA "\jl + cot 2 A = ^/cosec 2 A  1
J
Solution: cosA "\j 1 + cot A = cos A ^/cosec A = cosA . cosecA
C0S A . / 2l 7
= ^^ = cotA = ^cosec A  1
9 o j C — b
Example 6.11: If a sin 9 + b cos 9 = c, show that tan 9 = '
a — c
Solution:
2 2
a sin 9 + b cos 9 = c.
2 2 2
Dividing both sides by cos 9, we get a tan 9 + b = c sec 9
a tan 2 9 + b = c (1 + tan 2 9)
tan Q (a  c) = c  b
2n c ~b
:. tan 9 =
Example 6.12: that
1  cosA
1 + cosA
a — c
= cosec A  cotA
_ . . 1  cosA 1  cosA 1  cosA
Solution: consider, i + CO sA = 1 + cosA X T^A
194
(1cosA) 2 (\  cosAV
lcos 2 A V sinA
1  cosA 1  cosA 1 cosA
1 + cosA ~~ sinA ~~ sinA sinA
= cosec A  cotA
Example 6.13:
2 2 2 2
If x = a cos9 + b sin 9 and y = a sin 9  b cos 9, show that x +y = a + b
2 2 2 2
Solution: x + y = (a cos 9+ b sin9) + (a sin9  b cos 9)
2 2 2 2
= a cos 9 + b sin 9 + lab cos 9 sin9
2 2 2 2
+ a sin 9 + b cos 9  lab sin9 cos9
= a 2 (cos 2 9 + sin 2 9) + 6 2 (sin 2 9 + cos 2 9)
= a 2 + b 2
2 2
Example 6.14:Show that sin A.tanA+cos A . cotA+2 sinA . cosA=tanA + cotA
Solution: L.H.S. = sin A . » + cos A . • » + 2sinA cosA
3 3
sin A cos A _ .
= ^A +^A~ +2smA  cosA
_ sin 4 A + cos 4 A + 2sin 2 A . cos 2 A
sinA . cosA
(sin 2 A + cos 2 A) 2 1
sinA . cosA ~~ sinA . cosA
2 2
sin A + cos A r . o . o . , n
= sinA.cosA [vsm 2 A + cos 2 A=l]
2 2
sin A cos A
sinA cosA sinA cosA
citi A cosA
Hence the result . + ■ . = tanA + cotA = R.H.S.
Example 6.15: Show that 3(sinxcosx) +6(sinx + cosx) +4 (sin x+cos x) = 13
2
4 r 2~l 2 2 2
Solution: (sinx  cosx) = [(sinx  cosx) J = [sin x + cos x  2sinx cosx]
2
= [1 2sinxcosx]
2 2
= 1  4sinx cosx + 4sin x cos x (i)
2 2 2
(sinx + cosx) = sin x + cos x+ 2sinx . cosx
= 1 + 2sinx cosx (ii)
f\ f\ 9 % %
sin x + cos x = (sin x) + (cos x)
195
2 2 3 2 2 2 2
= (sin x + cos x)  3sin * . cos x (sin x + cos x)
2 2
= 1  3sin x cos x (iii)
Using (i), (ii) and (iii)L.H.S.
2 2
= 3(1  4sin x cos* + 4sin x . cos x)
2 2
+ 6(1 + 2sinx cosx) + 4(1  3 sin x cos x)
= 3+6+4
= 13 = R.H.S.
, , „ , _ , tan9 + sec9  1 1 + sin9
Example 6.16: Prove that — — r = 7~
tan9  sec9 + 1 cos9
Solution: L.H.S. =
tan9 + sec9  (sec 2 9  tan 2 9)
tan9  sec9 + 1
tan9 + sec9  (sec9 + tan9) (sec9  tan9)
tan9  sec9 + 1
(tan9 + sec9) (1  sec9 + tan9)
= — — — = tan9 + sec9
(tan9  sec9 + 1)
sin9 1 _ sin9 +1 _
cos9 cos9 ~~ cos9
EXERCISE 6.3
(1) Prove the following:
(i) sin A  cos A = 1  2cos A
(ii) sin 3 A  cos 3 A = (sinA  cos A) (1 + sinA cosA)
(iii) (sin9 + cos9) 2 + (sin9  cos9) 2 = 2
(iv) (tan9 + cot9) = sec 9 + cosec 9
,  1 1 2^ , n secx + tanx , N ?
(v) ', r~r + r~z = 2sec 9 (vi) = (secx + tanx)
1 + sin9 1  sin9 secx  tanx
(ix)
cosec 9 „ 1
— — r = cos9
+ tan9
1 1 + cos9
(vii) — TK — I — 7: = cos9 (viii) : — „ r = sec9  tan9
v ' cot9 + tan9 tan9 + sec9
cosec9  cot9 sin9
(x) (sec9 + cos9) (sec9  cos9) = tan 9 + sin 9
(2) If tan9 + sec9 = x, show that 2tan9 = x  — , 2sec9 = x + —
Hence show that sin 9 = ~
x 2 l
x"+l
196
2 2 l —
(3) If tan9 + sin6 = p, tan8  sin6 = q and p > q then show that p q = 4^jpq
sccA cosccA
(4) Prove that (1 + cotA + tanA) (sinA  cos A) = y~ 5 —
cosec A sec A
,_ _ , cosA sinA .
(5) Prove that  — ; — r + — = sin A + cos A
' 1  tanA 1  cotA
(6) Prove the following :
/ 1 + sinA A A ,_ /1+cosA
W "\ / 1 ""a" = sec ^ + tatL ^ ( u ) ^ ' '• ~ = cosec A + cotA
1  sinA
1  cosA
(sinA * 1)
(cosA * 1)
/1sine
(111) \ I \ . . ~ = sec9  tan9
1 + sin9
(7) If cos9 + sin9 = \[2 cos9, show that cos9  sin9 = y2 sin9
(8) Prove that (1 + tanA + sec A) (1 + cotA  cosec A) = 2
6.3 Compound Angles
6.3.1 Compound Angles A + B and A  B
In the previous chapter we have found the trigonometrical ratios of angles
such as 90°+ 9, 180° + 9, ... which involves single angle only. In this chapter
we shall express the trigonometrical ratios of compound angles such as A + B,
A  B, ... interms of trigonometrical ratios of A, B, ....
It is important to note that the relation/(x + y) =f(x) +f(y) is not true for all
functions of a real variable. As an example all the six trigonometrical ratios do
not satisfy the above relation.
cos (A + B) is not equal to cosA + cosB.
Let us develop the identity
cos(A  B) = cosA cosB + sinA sinB
ftp*
Q(ix*iB,anB)
Fig. 6. 18
Fig. 6. 19
197
Let P and Q be any two points on the unit circle such that XOP = A and
XOQ = B. Then the coordinates of P and Q are (cos A, sinA) and
(cosB, sinB) respectively.
PQ 2 = (cosA  cosB) 2 + (sinA  sinB) 2
= (cos A  2cosA cosB + cos B) + (sin A  2sinA sinB + sin B)
= (cos 2 A + sin 2 A) + (cos 2 B + sin 2 B)  2cosA cosB  2sinAcosB
= 1+1  2 cos A cosB  2 sinA sinB=2  2 (cosA cosB+sinA sinB) ... (1)
Now imagine that the unit circle above is rotated so that the point Q is at
(1, 0). The length PQ has not changed.
PQ 2 = [cos (A  B)  l] 2 + [sin (A  B)  0] 2
= [cos 2 (A  B)  2cos(A  B) + l] + sin 2 (A  B)
= [cos 2 (A  B) + sin 2 (A  B)] + 1  2cos (A B) = 1 + 1  2cos(A  B)
= 22cos(AB) ...(2)
From (1) and (2), 2  2cos(A  B) = 2  2 (cosAcosB + sinA sinB)
=> cos(A  B) = cosA cosB + sinA sinB
Next let us consider cos(A + B). This is equal to cos [A  ( B)] and by
cosine of a difference identity, we have the following:
cos(A + B) = cosA cos( B) + sinA . sin ( B)
But cos( B) = cosB and sin(B) =  sin B
.'. cos(A + B) = cosA cosB  sinA sinB.
To develop an identity for sin(A + B), we recall the following:
(n
sin9 = cos I 2~ 9
In this identity we shall substitute A + B for
sin (A + B) = cos
f(A + B)
= cos
fAlB
We can now use the identity for the cosine of a difference.
= cosl2"Aj . cosB + sin I 2~ A  .sinB
= sinA . cosB + cosA . sinB
Thus, sin (A + B) = sinA . cosB + cosA sinB
198
To find an identity for the sine of a difference, we can use the identity just
derived, substituting  B for B
sin (A  B) = sin [A + ( B)]
= sinA cos( B) + cosA . sin( B)
sin(A  B) = sinA cosB  cosA sinB
An identity for the tangent of a sum can be derived using identities already
established.
, A _ sin (A + B)
tan(A + B) = cos(A + B)
sinA . cosB + cosA sinB
cosA cosB  sinA sinB
Divide both Numerator and Denominator by cosA cosB
sinA cosB cosA sinB
cosA cosB cosA cosB
 cosA cosB sinA sinB
cosA cosB cosA cosB
tanA + tanB
tan (A + B) =  — ; — : — : — —
v ' 1  tanA . tanB
Similarly, an identity for the tangent of a difference can be established.
T ■ ■ , , . t^ tan A  tan B
It is given by tan (A  B) =  — : — : — : — ~
b J ■ ' 1 + tanA . tan B
(1) sin (A + B) = sinA cosB + cos A sinB
(2) sin (A  B) = sinA cosB  cosA sinB
(3) cos (A + B) = cosA cosB  sinA sinB
(4) cos (A  B) = cosA cosB + sinA sinB
tanA + tanB
(5) tan (A + B) =
1  tanA tanB
tanA  tanB
(6) tan ( A  B )=l + tanA.tanB
Example 6.i7:Find the values of (i)cos 15° (ii)cos 105° (iii)sin 75° (iv)tan 15°
Solution:
(i) cosl5° = cos(45°30°) = cos45° cos30° + sin45° sin30°
_J_ i/l 1 1 a/3+1 a/3 + 1 a/2 \/6+\/2
"f2 2 + ^2 2 ^  2f2 ^ " 4
(ii) cosl05° = cos(60° + 45°) = cos60° cos45°  sin60° sin45°
199
_I J_ _i/l J_ _ l>/3 a/2\/6
2^/2 2^2^ 4
(iii) sin75° = sin(45° + 30°) = sin45° cos30° + cos45° sin30°
_J_ i/1 J_ 1 a/3+1 a/6 + a/2
>/2 • 2 + ^2 2 2a /2  4
„,„ ™ tan45°  tan30° \/3
(iv) tanl5° = tan(45°  30°) = tan450 = — — p
1+1 Vf
3\/3
= 2a/3
3+a/3
A, B are
Solution: cos (A + B) = cosA cosB  sinA sinB
3 12
Example 6.18: If A, B are acute angles, sinA = t ;cos B= tt , find cos(A + B)
H
cosA = yj 1  sin A = \
4
 5
X/ 1 169
33
_ 65
sinB = "n/I cos 2 B = \
/A „, 4 12 3 5
. cos(A + B) = 5 . 13  5 . 13
5
 13
Example 6.19: Show that (i) sin(A + B) sin(A  B) = sin A  sin B
(ii) cos(A + B) cos(A  B) = cos 2 A  sin 2 B
sin(A +B) sin(A  B) = (sin A cosB + cosA sinB) (sinA cosB  cosA sinB)
2 2 2 2
= sin A cos B  cos A sin B
= sin 2 A (1  sin 2 B)  (1  sin 2 A) sin 2 B
= sin A  sin B
cos(A + B) cos(A  B) = (cosA cosB  sinA sinB) (cosA cosB + sinA sinB)
2 2 2 2
= cos A cos B  sin A sin B
= cos 2 A (1  sin 2 B)  (1  cos 2 A) sin 2 B
2 2
= cos A  sin B
Example 6.20: If A + B = 45°, show that (1 + tanA) (1 + tanB) = 2 and hence
deduce the value of tan 22 ^
Solution: Given A + B = 45° => tan (A + B) = tan45°
200
tanA + tanB
1  tanA . tanB
i.e. tanA + tan B = 1  tanA . tanB
i.e. 1 + tanA + tanB = 2  tanA tanB (add 1 on both sides)
1 + tanA + tanB + tanA tanB = 2
i.e. (1 +tanA)(l + tanB) = 2
Take A = B then 2A = 45° => A = 22 ^ = B
1 + tan 22 2 J =2 => 1 + tan 22 ^ = ± V 2
.. tan 22 2 = ±yJ2  1
Since 22 ~ is acute, tan 22 7 is positive and therefore tan 22 7 = "v2  1
Example 6.21:
tan69° + tan66° . . tan (A  B) + tanB
(i) Prove that  — : — ,„„ : — — =  1 (n)  — : — — — 7— — 7 = tan A
w 1  tan69° tan66 1  tan(A  B) tanB
cos 17° + sinl7
Solution:
tan69° + tan66° , rno rro ^
(1 ) 1 _ tan69 o tan66 o=tan(69° + 66°)
(iii) T^Z • no = tan62°
coslv sinl7
= tan (135°) = tan (90° + 45°) =  cot45° =  1
,.. N tan (A  B) + tanB . „ N ,
(11) i — t 7T DW D = tan [(A  B) + B] = tanA
1  tan(A  B) tanB LV ' J
cos 17° + sinl7°
(iii) L.H.S. =
cosl7°sinl7°
Divide both Numerator and Denominator by cosl7°
T TTO l+tanl7° tan 45° + tanl7° , _ „
LHS " = T^nT^ = ltan45°tanl7° ( " tan45 = 1}
= tan (45° + 17°) = tan62° = R.H.S.
r n „ 1 In
Example 6.22: Prove that (i) tan I 7 + 9 I tan 1 7  9 J = 1
1 7t
(ii) If tanA = 3 and tanB = 7 , prove that A  B = 7
201
Solution:
(i) L.H.S. = tan (7 + 9 J tan (78
1 + tanjn f 1  tangj , ( 71 ,
r^nej Irr^neJ = ! l vtan 4 =1
31 $
tanA  tanB 2 2 7:
(ii) tan(AB)= 1+tanA tanB = — ^ =7 =1 = tan
1+3.2 2
71 71
tan (A  B) = tan 7 => A  B = 7
4 5
Example 6.23: If cos(a + (3) = 7 and sin (a  (3) = tt where (a + P) and
(a  (3) are acute, find tan 2a.
Solution:
4 3
cos (a + P) = 7 => tan (a + P) = 7
sin (a  P) = tt => tan (a  p) = 77
2a = (a + P) + (a  P)
.. tan2a = tan [(a + P) + (a  P)]
3 JL H
tan (a + p) + tan(g  p) 4 + 12 _12 _ 56
~~ 1  tan(a + (3) . tan(a p) _ , 3J_ _ jJ_~33
1_ 4 X 12 16
Example 6.24: Prove that tan3A  tan2A  tanA = tanA tan2A tan3A
Solution:
„ ^ tanA + tan2A
tan3A = tan(A + 2A) =  — ; — — — 77
v ' 1  tanA tan2A
i.e. tan 3 A (1  tanA tan2A) = tanA + tan2A
i.e. tan3A  tanA tan2A tan3A = tanA + tan2A
.. tan3A  tan2A  tanA = tanA tan2A tan3A
EXERCISE 6.4
(1) Find the values of (i) sin 15° (ii) cos 75° (iii) tan 75° (iv) sin 105°
(2) Prove that
(i) sin (45°+A) = —f= (sinA+cosA) (ii) cos(A+45°) = —fe (cosAsinA)
202
(3) Prove that
(i) sin (45° + A)  cos(45° + A) = ^2 sinA
(ii) sin(30° + A) + sin(30°  A) = cosA
(4) Prove that (i) cos(A + B) cos(A  B) = cos 2 B  sin 2 A
(ii) sin(A + B) sin(A  B) = cos B  cos A
(5) Prove that cos 2 15° + cos 2 45° + cos 2 75° = ^
(6) Prove that (i) sinA + sin(120° + A) + sin(240° + A) =
(ii) cos A + cos(120° + A) + cos(120°  A) =
(7) Show that
(i) cosl5°sinl5°=7^ (ii) tanl5°+cotl5°=4 (iii) cot 75°+tan75° = 4
(8) (i) Find sin45° + sin30° and compare with sin 75°
(ii) Find cos45°  cos30° and compare with cosl5°.
(9) Show that
(i) tan70° = 2 tan50° + tan 20°
(ii) tan72° = tan 18° + 2tan54° (Hint : tanA tanB = 1 if A + B = 90°)
cos 1 1 ° + sin 1 1 ° cos29° + sin29°
(111) cosll°sinll° = tan56 (1V) cos29°sin29° = tan 74
nm p tv, t sin ( A ~ B ) A sin(BC) A sin (C  A)
(10) Prove that sinA sinB + sin B sinC + sin C sin A =0
5 1
(1 1) ( i) If tanA = g , tan B = yy show that A + B = 45°
1 1 %
(ii) If tan a = y and tan P = t , show that a + (3 = t
(12) If A + B = 45°, show that (cotA1) (cotB  1) = 2 and deduce the value
of cot 22 ^
(13) If A + B + C = 7i, prove that
(i) tanA + tanB + tanC = tanA tanB tanC
(ii) tan2A + tan2B + tan2C = tan2A tan2B tan2C
(14) If sinA = ~ , sinB = t find sin (A + B), where A and B are acute.
(15) Prove that (i) sin (A + 60°) + sin(A  60°) = sinA
(ii) tan4A tan3A tanA + tan3A + tanA  tan4A =
203
6.3.2 Multiple angle identities:
Identities involving sin2A, cos2A, tan3A etc. are called multiple angle
identities. To develop these identities we shall use sum identities from the
preceding lesson.
We first develop an identity for sin2A.
Consider sin (A + B) = sinA cosB + cosA sinB and put B = A
sin2A = sin (A + A) = sinA cosA + cosA sinA
= 2 sinA cosA
Thus we have the identity  sin2A = 2sinA . cosA
Identities involving cos2A and tan2A can be derived in much the same
way as the identity above
cos2A = cos (A + A) = cosA cosA  sinA sinA
sin A
cos2A = cos A
Thus we have the identity
Similarly we can derive
The other useful identities for cos2A can easily be derived as follows:
cos2A = cos 2 A  sin 2 A = (1  sin 2 A)  sin 2 A
y
2 2
cos2A = cos A  sin A
2tanA
La
tizA  t
1  tan 2 A
= 1
2sin 2 A
cos2A = cos A  sin A = cos A  (1  cos A)
= 2cos 2 A  1
From
cos2A = 1
sin A = —
2sin A, also we have
■ cos2A
Also,
Hence
cos2A = 2cos A  1
2 1 + cos2A
cos A = ~
cos2A
2, 1
tan A =
1 + cos2A
sin2A = 2sinA cosA
2 sinA 7 2 tanA
cos A = ■
cosA
sec 2 A
2tanA
1 + tan 2 A
204
cos2A = cos A  sin A = cos A 1 y~
V cos A
= cos 2 A(l tan 2 A)
1  tan 2 A 1  tan 2 A
sec 2 A 1 + tan 2 A
Thus we have sin2A = 2 sinA . cosA
2 2
cos2A = cos A  sin A
cos2A = 1  2sin A
cos2A =
2 cos 2 A 
tan2A =
2tanA
1  tan 2 A
sin2A =
2tanA
1 + tan 2 A
cos2A =
1  tan 2 A
1 + tan 2 A
.3.3:
Trigonometrical ratios of
atios
ofy
sinA = sinl 2 x y
=
2 sin y . cos y
cos A =
cos 12 xy 1 =
=
o 2 A i
2 cos y  1
=
1  2 sin y
tan A =
tan ( 2 x y J
2 tany
—
2 A
1  tan y
2 A . 2 A
s y  sin y
205
Similarly, we can prove the following identities
A
sinA =
cosA =
2 tan "~
1 + tan y
2 A
1  tan y
i 2A
1 + tan y
. 2 A 1  cosA
in 2=~
2 A 1 + cosA
cos 2  2
2 A _ 1  cosA
tan 2 = 1 + cosA
A sinA A 1  cosA
Also note that tany= 1+cosA and tan y= sinA
3
Example 6.25: If sin9 = 77 and 9 is acute, find sin29 ?
Solution: sin9 = g ; cos 9 = \j 1  sin 9 = A/l^J = s~~
sin 29 = 2 sin9 cos9 = 2 . o . o  ,,
Example 6.26: Find (i) sinl5° (ii) tanl5°
64~8
3 ^/55 3a/55
o , . ,^ . ,„ • 39° / lcos39
Solution : (l) sin 15 = sin^  = v 9
(ii) tan 15° = tan y
sin30°
6.3.4 Trigonometrical ratios involving 3A
sin3A = sin (2A + A) = sin2A . cosA + cos2A . sinA
= 2 sinA cos 2 A + (1  2sin 2 A) sinA
= 2sinA (1  sin 2 A) + (1  2sin 2 A) sinA
206
= 3sinA  4sin 3 A
Similarly, cos3A = 4 cos A  3cosA
tan2A + tanA
tan3A = tan (2A + A) =  — ; — rr— ; — 
■ ' 1  tan2A . tanA
2 tanA "\
^— + tanA
1  tan 2 Ayl
2tanA
1  tanA . ^~
1  tan 2 A
_ 3 tanA  tan 3 A
1  3tan 2 A
Example 6.27: Prove that cos A  sin A = cos2A
Solution:
L.H.S.
= 1 . cos2A = cos2A = R.H.S.
L.H.S. = (cos 2 A + sin 2 A) (cos 2 A  sin 2 A)
Example 6.28:
Solution:
cot 3 A  3cotA
that cot3A = ,
3cot 2 A  1
1 3
„ cot 3 A3cotA tan 3 A tanA
R.H.S. — o — i
3cot 2 A  1 3 j
1  3tan 2 A
3tanA  tan 3 A
tan 2 A
= —krr = cot3A = L.H.S.
Example 6.29:
1  cosB
If tanA = — jg — , prove that tan2A = tanB, where A and B are acute
angles.
„ ■ 2B . B
, _ 2sin Tr sinTr n
1  cosB 2 2 B
Solution : R.H.S = sinB = g g" = g = tan ^
2sin 2 ■ cos 2 cos 7
B
.". tan y = tanA
207
B
^A=2" =^> B = 2A
Therefore tan2A = tanB
Example 6.30: Show that 4 sinA sin (60° + A) . sin (60°  A) = sin3A
Solution: L.H.S. = 4 sinA sin (60° + A) . sin (60°  A)
= 4sinA { sin (60° + A) . sin (60°  A)}
= 4sinA {sin 2 60  sin 2 A}
= 4sinA It  sin 2 Af = 3sinA  4sin 3 A = sin3A
= R.H.S.
Example 6.31: Prove that cos20° cos40° cos80°=
L.H.S. = cos20° cos40° cos80°
= cos20° {cos (60°  20°) cos(60° + 20°)}
= cos20° [cos 2 60°  sin 2 20°]
= cos20° t  sin 2 20°
= \ cos20° { 1  4(1  cos 2 20°) }
= T {4cos 3 20°  3 cos20°} = 4 [cos3 x 20°]
= 4 xcos60° = g = R.H.S.
Example 6.32: Find the values of:
(i)sinl8° (ii)cosl8° (iii) cos36° (iv) sin36° (v) sin54° (vi) cos54°
Solution:
(i) Let 9=18° then 59 = 90° => 29 = 90°  39
=> sin29 = sin(90°  39) = cos39
=^> 2sin9 cos9 = 4cos 3 9  3cos9
=> 2sin9 = 4cos 2 9  3 (.cos 9*0)
=> 2sin9 = 1  4sin 2 9
=> 4sin 2 9 + 2sin9 1=0
208
sin9 =
2+V4+16  1 ±\/5
since sin 18° is positive, sinl8° = ,
(ii) cosl8° = V 1  sin z 18 = A / 1  P^
(iii) cos36° = 1  2sin 2 18° = 4^
(iv) sin36° = Vlcos 2 36° = ^^
(v) sin54° = sin (90°  36°) = cos36° = 4 —
(vi) cos54° = cos (90°  36°) = sin36° = ^ — j^~
EXERCISE 6.5
(1) Prove the following:
1 n n 1
(i) 2sinl5° cosl5° = 2 (ii)sing cos 77 = 7—7^
(iii) sin72° = ^±M ( . v) cos72o = ^5_±
, , 2 tan22 j°
(v)l2sin 2 22f=j= (vi) — p = 1
I  tan 22^
71
(2) Show that 8 cos 75  6cos 77 = 1
n
(3) If tan 2 = (2  a/3) find the value of sin9
^ N „ , 1 + sin 9 cos 9 9
(4) Prove that : — 7 zr = tan x
1 + sin 9+ cos 9 2
(5) Prove that
(i) cos 2 I 4  9 1  sin 2 (7  9 1 = sin29 (ii) sec29+tan29 = tan [4 + 9
3
(6) (i) If tan9 = 3 find tan39 (ii) If sinA = 5 find sin3A
209
1 1 71
(7) If tana = t and tan (3 = y show that 2a + (3 = t
(8) If 2 cos9 = x + — then prove that cos29 = 9 •* + ~
6.3.5 Transformation of a product into a sum or difference
We know that
sin(A + B) = sinA cosB + cos A sinB ... (1)
and sin(A  B) = sinA cosB  cosA sinB ... (2)
Adding (1) and (2), we get
sin(A + B) + sin (A  B) = 2 sinA cosB ... (I)
Subtracting (2) from (1)
sin(A + B)  sin(A  B) = 2 cosA sinB . . . (II)
Again
cos(A + B) = cosA cosB  sinA sinB ... (3)
cos(A  B) = cosA cosB + sinA sinB ... (4)
(3) + (4) => cos(A + B) + cos(A  B) = 2 cosA cosB . . . (Ill)
(4)  (3) cos(A + B)  cos(A  B) =  2sinA sinB . . . (IV)
Now, let A + B = C and A  B = D then
C+D CD
2A = C + D (OR) A = —j— and 2B = C  D (OR) B = — ^—
Putting these values of A and B in the above
four formulae I, II, III and IV, we get
,, ■ ~ •  ~ • C + D CD
1) sinC + sinD = 2 sin — ~ — • cos — 7 —
„x ■ „ ■ ~ „ C + D . CD
2) sinC  sinD = 2 cos — ~ — • sm — 7 —
r. „ C + D CD
3) cos C + cosD = 2 cos — ~ — • cos — 9 —
C+D CD
4) cosD  Cos C = 2sin — ~ — • s i n — 7 —
Example 6.33: Express as sum or difference of following expressions,
(i) 2sin29 . cos9 (ii) 2 cos29 cos9 (iii) 2 sin 3A . sinA
3A 5A
(iv) cos79.cos59 (v)cos j cos ~y~ (vi) cos39.sin29 (vii) 2cos3A . sin5A
210
Solution:
(i) 2 sin 29 . cos9 = sin(29 + 9) + sin (29  9) = sin39 + sin9
(ii) 2 cos29 . cos9 = cos(29 + 9) + cos (29  9) = cos39 + cos9
(iii) 2 sin3A . sinA = cos(3A  A)  cos(3A + A) = cos2A  cos4A
(iv) cos79 . cos59 = \ [cos(79 + 59) + cos(79  59)] = ^ [cos!29 + cos29]
3A 5A 1
(v) cos —j . cos^  = 2
3A 5A^ f3A 5A
COS I ~y~ + ~7T I + COS i j ~ 1
= j [cos4A + cos(A)] = 2 [cos4A + cosA]
(vi) cos39 . sin29 =j [sin (39 + 29) sin(39  29)] = j [sin 59  sin 9]
(vii) 2 cos3A . sin5A = sin(3A + 5A)  sin(3A  5A) = sin8A  sin(2A)
= sin8A + sin2A
Example 6.34: Express the following in the form of a product:
(i) sin4A + sin2A (ii) sin5A  sin3A (iii) cos3A + cos7A
(iv) cos2A  cos4A (v) cos60°  cos20° (vi) cos55° + sin55°
Solution:
• „ A • „ . „ ■ f4A + 2A\ f 4A  lK\
(I) sin4A + sin2A = 2 sin I ~ J cos I 9 1 = 2sin3A cosA
■ *. • o« „ r5A + 3A^ . f5A  3A^ „ A A . A
(II) sin5A  sin3A = 2cos I ~ J sin I 9 J = 2 cos4A sinA
n. „ f3A + 7A\ (3 A  7 A
(III) cos3A + cos7A = 2 cos I ~ 1 cos I ir~
= 2cos5A cos (2A)=2cos5A cos2A
„ • f2A + 4M . f2A  4A^
(iv) cos2A  cos4A =  2sin I ~ 1 sin I ~ 1
= 2sin3A sin(A)=2sin3A sinA
(60° + 20°) f60°  20°^
(v) cos60°  cos20° =  2 sin i ^ sin I 2 J = " 2sin40 ° sin20 °
(vi) cos55° + sin55°=cos55° + cos(90°55°) = cos55° + cos35°
55° + 35° 55°  35° „ ien
= 2 cos ^ cos ^ = 2cos45 coslO
= 27= cosl0°=^/2 coslO
211
a/3
Example 6.35: Show that sin 20° sin40° sin80°=
Solution:
L.H.S. = sin20 sin40° sin80°= sin20° ^ {cos40° cosl20 c
1 __. f ._. 1
= 2 sin20° cos40° + 2
= 2 sin20° cos40° + 4 sin20°
= 4 (sin60°  sin20°) + 4 sin20° = 4 sin60°
= f =R.H.S.
Example 6.36: Prove that 4(cos6° + sin24°) = ^3 + y/B
Solution:
4(cos6° + sin24°) = 4 (sin84° + sin24°) [v cos6° = cos(90°84) = sin84°]
„ „ . (84° + 24°^ f84°  24°'
= 4 . 2sin ~ cos
2 ; wo v 2
V5 + 1
= 8 sin54° . cos30° = 8
= >/l5 +^3
Example 6.37:
Prove that (i) cos20°+cosl00° + cosl40° = (ii) sin50 o sin70°+sinl0°=
Solution:
(i) L.H.S. = cos20° + (cosl00° + cosl40°)
™ n „ ( lOQ + 140°^ flOO140' "
= cos20 + 2cos I 2 1 • cos I 7
= cos20° + 2cosl20° cos( 20°) = cos20° + 2[ 2) cos20°
= cos20°  cos20° = = R.H.S.
(ii) L.H.S. = sin50°  sin70° + sinl0°
f50 + 70°^ . f50  70°^ . ,„ n
= 2 cos ~ .sin ~ + sinlO
212
= 2 cos60° . sin( 10°) + sinl0° = 2x^ ( sinl0°) + sinl0°
=  sinl0° + sinl0° = = R.H.S.
6.3.6 Conditional Identities
Example 6.38:
If A + B + C = 7i, prove that sin2A + sin2B + sin2C = 4sinA sinB sinC
Solution:
L.H.S. = sin2A + sin2B + sin2C = (sin2A + sin2B) + sin2C
= 2sin(A + B) cos(A  B) + sin2C
= 2sin(Tt  C) cos(A  B) + sin2C
= 2sinC cos(A  B) + 2sinC cosC
= 2sinC {cos(A  B) + cosC}
I
= 2 sinC lcos(A  B) + cos(180  A + B)J
= 2 sinC {cos(A  B)  cos(A + B)} = 2sinC {2 sinA sinB}
= 4 sinA sinB sinC = R.H.S.
Example 6.39:
If A + B + C = 180° Prove that cos2A + cos2Bcos2C = l4sinA sinB cosC
Solution:
L.H.S. = cos2A + (cos2B  cos2C)
= 1  2sin 2 A + { "2 sin(B + C) sin(B  C) }
= 1  2sin 2 A  2sin(180°  A) sin(B  C)
= 1  2sin 2 A  2sinA sin(B  C)
= 12 sinA [sinA + sin (B  Q]
= 1  2sinA [sin(B + C) + sin(B  C)] , [v A = 180°  (B + C)]
= 1  2sinA [2sinB cosC]
= 1  4sinA sinB cosC = R.H.S.
Example 6.40:
If A+B+C = %, prove that cos 2 A + cos 2 B  cos 2 C = 1  2sinA sinB cosC
Solution:
L.H.S. = cos 2 A + cos 2 B  cos 2 C = (1  sin 2 A) + cos 2 B  cos 2 C
= 1 + (cos 2 B  sin 2 A)  cos 2 C
= 1 + cos(A + B) . cos(A  B)  cos 2 C
= 1 + cos(tt  C) cos(A  B)  cos C
= 1  cosC. cos(A  B)  cos 2 C
213
= 1  cosC [cos(A  B) + cosC]
= 1  cosC [cos(A  B) cos (A + B)] = 1  cosC [2sin A sinB]
= 1  2sinA sinB cosC = R.H.S.
EXERCISE 6.6
(1) Express in the form of a sum or difference:
(i) 2sin49 cos29 (ii) 2cos89 cos69 (iii) 2cos79 sin39
(iv) 2sin3A sinA (v) 2cos6A sin3A (vi) cos49 sin99
3A A 7A 5A 59 49
(vii) cos ~y sin y (viii) sin~y~ cos ~y (ix) cos y cos y
(2) Express in the form of a product:
(i) sinl3A + sin5A (ii) sinl3A  sin5A (iii) cosl3A + cos5A
(iv)cosl3Acos5A (v) sin52°  sin32° (vi) cos 51° + cos23°
(vii) sin80°  cos70° (viii) sin50° + cos80° (ix) sin20° + cos50°
(x) cos35° + sin72°
3
(3) Prove that sin20° sin40° sin60° sin80°=Yg
(4) Prove that cos20° cos40° cos60° cos80 o = yg
(5) Prove that sin50°  sin70° + cos80° =
? 9 9 [ Ct + 6
(6) Prove that (cosa + cos(3) + (sina  sin(3) =4cos I — y^
,„ _ , _ sin3AsinA _ A ,.. cos2A cos3A A
(7) Prove that (l) : r— = cot2A (n) •  . , — •  . = tan tt
' ' cosA  cos3A v ' sin2A + sin3A 2
(8) A + B + C = 7t, prove that sin2A  sin2B + sin2C = 4 cosA sinB cosC
(9) If A + B + C=180°,
2 2 2
prove that sin A + sin B + sin C = 2 + 2cosA cosB cosC
A B B C C A
(10) If A+B+C = 7i, prove that tan y tan y +tan y tan y +tan y tan y = 1
„.„, T „ . „ „„„ , , sin2A + sin2B + sin2C
(11) If A + B + C = 90°, show that — —. r^r~ — = cot A cot B
sin2A + sin2B  sin2C
2 A 2 B 2C
(12) Prove that A + B + C = 71, prove that sin y + sin y + sin y
1 o • A • B • C
= 12 siny sin y sin y
214
6.4 Trigonometrical Equations
An equation involving trigonometrical function is called a trigonometrical
equation.
1 2 1
cos9 = ;y> tan 9 = 0' cos 9 ~~ 2sin9 = y are some examples for
trigonometrical equations. To solve these equations we find all replacements for
the variable 9 that make the equations true.
A solution of a trigonometrical equation is the value of the unknown angle
that satisfies the equation. A trigonometrical equation may have infinite number
of solutions. The solution in which the absolute value of the angle is the least is
called principal solution. Note that trigonometrical equations are different
from trigonometrical identities. It is possible that some equations may not have
solution. For example cos9 = 4 has no solution. The expression involving
integer V which gives all solutions of a trigonometrical equation is called the
general solution.
6.4.1 General solutions of sin 9 = ; cos9 = ; tan 9 =
Consider the unit circle with centre at O(0, 0)
Let a revolving line OP, starting from OX, trace XOP =9 Draw PM
perpendicular to OX
(l)sin9 = tY
In the right angled triangle OMP we have OP = 1
unit,
MP
sin9 =
OP
sin9 = MP
If sin9 = 0, then MP = 0, i.e. OP coincides with
OX or OX'
V
Fig. 6.20
.'. XOP = 9 = 0, 7t, 2tc, 3tc, ... [in the anti clockwise direction]
or 9 =  7i,  2tc,  3tc, [in the clockwise direction]
i.e. 9 = or any + ve or  ve integral multiple of n.
Hence the general solution of sin9 = is given by 9 = nn, n e Z,
where Z is the set of all integers.
(2) cos9 =0
In the right angled triangle OMP we have cos9 = ~Qt7 = OM (.OP = 1 unit)
215
If cosG = 0, then OM =
i.e. OP coincides with OY or OY'
i.e. 9 = 2 > "t~ > 5y , [in anticlockwise direction)
. tc 3tc 5n .
or 9 =  2 » _ "t" . _ "t" [m clockwise direction]
solution which is in
71 71 1
for sine, in I  y, 9" J f° r tangent and in [0, Tt] for
i.e. 9 = + I odd multiple of ~
Hence the general value of 9 is given by 9 = (2« +1)9" ,neZ
(3) tan9 =
MP
In the right angled triangle OMP, if tan9 = then q^ = or MP =
Proceeding as in (1), we get 9 = tin, n e Z
Thus, (1) If sin9 = 0, 9 = tin, neZ
(2) If cos9 = 0, 9 = (2«+l) , weZ
(3)Iftan9 = 0, 9 = mtc, weZ
When a trigonometrical equation is solved, among all solutions the
71 71
2 ' 2
cosine, are the principal values of those functions.
Example 6.41:
Find the principal value of the following :
(i) cosx = 2 (ii) cos 9 =  2 (***) cosec9 =  "7^
(iv) cot9 =  1 (v) tan9 = ^3
Solution: (i) cos* = 2 >
.". x lies in the first or fourth quadrant. Principal value of x must be in
[0, Tt]. Since cos* is positive the principal value is in the first quadrant
cosx = 2 = cos a an ° a e [0> TC ]
n
:. The principal value of x is t .
216
a/3
(ii) cos9 = ^ <0
Since cos 9 is negative, 9 lies in the second or third quadrant. But the
principal value must be in [0, n] i.e. in first or second quadrant. The principal
value is in the second quadrant.
.. cos9 =  ^r = cos (180°  30°) = cosl50°.
5n
The principal value is 9 = 150° = ~~z~ .
(iii) cosec9 =  ~~7^ => sin9 =  2 <
.". 9 lies in the third or fourth quadrant. But principal value must be in
n %
2' 2_
n
i.e. in first or fourth quadrant. .\ 9 =  t
(iv) cot9 =  1 .. tan9 =  1 <
( n n
:. 9 is in the second or fourth quadrant. Principal value of 9 is in I  t", 9
.". the solution is in the fourth quadrant.
f n] . . n ( n n
cot {4) =~ l => e = "4 e l"2' 2
6.4.2 General solutions of sin 9 = sin a ; cos 9 = cos a ; tan 9 = tan a
(1) sin9 = sina y ^a< y ie  a
^> sin9  sina =
'2' 2
9 + a] . fQa
2 cos 1 — j — j . sin 1 — ^ — I =
9 + a\ „ .fQa
cos I — ^ — I =0 or sin I — ~" — I =
9 + a 7t 9  a
— ~ — = (2m + 1) y , or — j — = tin, n e Z
9 + a = odd multiple of n or 9  a = even multiple of n
9 = (odd multiple of n)  a . . . (1)
or 9 = (even multiple of n) + a . . .(2)
217
Combining (1) and (2), we have
9 = tin + ( l)".a, where n e Z
(2) cos9 = cosa < a < n i.e. a e [0, n]
=> cos9  cosa =
9 + a\ . (Qa
■ 2sin I — j — I • sin I — n — I =
9 + a\ „ . (Q  a
sin I — 2 — I = or sin[ — ^ —  =
9 + a „ 9  a „
=> — j — = tin ; n€ Z or — ~ — = tin ; n e Z
=> 9 = 2wtc  a or 9 = 2«7T + a
Hence Q = 2nn±a ' n eZ.
(3) tan9 = tana  y < a < t i.e. a e I  x, 7
sin9 sina
cos9 _ cosa
=> sin9 cosa  cos9 sina =
=> sin (9  a) =
=> 9  a = tin, n e Z
=> 9 = wrc + a, « eZ
Thus, we have sin 9 = sina =>9 = ra + (lfa;/i6Z
cos 9 = cos a => 9 = 2wtc + a ; n e Z
tan 9 = tan a => 9 = wTC + a;weZ
Example 6.42: Find the general solution of the following :
(i) sin9 = 2 (ii) sec9 =  ^2 (iii) cos 2 9 = 4 (iv) cot 2 9 = 3 (v) sec 2 9 = j
Solution: (i) sin9 = «
1 tc n
sin 9 = 9 = sin t which is of the form sin9 = sina where a = ~z
n
.'. The general solution is 9 = tin + ( 1)" . ~z ', n e Z
(ii) sec9 = ^J2 =^> cos 9 =  75 <
Principal value of 9 lies in [0, n]
As cos 9 is negative, the principal value of 9 lies in second quadrant.
218
3n f it] n I
COS "J" = COS I 7T  4 I =  COS 4 =  ~jE
:. 9 = 2nit ± ~r ; n e Z
(iii) We know that cos29 = 2cos 91
A . 1 , 1 it 2it
= 2 It I 1=t1=t= cos t = cos :
2u
.". 29 = 2mt + ~t~ ;neZ
9 = «jr + T ; n e Z
22 2
(iv) We know that 1 + cot 9 = cosec 9 => 1 + 3 = cosec 9
2 2 1
.". cosec 9 = 4 or sin 9 = T
cos29 = 12 sin 9=1 2l tJ = t = cos t
71
.". 29 = 2mt ± t ; « e Z
9 = Mjr + T ;weZ
2 2 4 1
(v) We know that tan z 9 = sec z 9 1 = t 1=3
ltan 2 9
cos 29 = t =
1 + tan z 9
.I
.4
2
3
_ 4
3
1
~2
COS 29 = T = COS T
29 = 2nit ± t ; neZ
9 = M7r + T ; n e Z
Note : Solve : sin9 = '
^
7T 4rt
There are two solutions in < 9 < 2% i.e. 9 =  t and ~t~
219
The general solution is
71
9 = wi + (l) n lTj ; n eZ ...(1)
Even if we take 9 = n n + ( 1)" I ~r~ J ;neZ . . . (2)
The solution will be the same although these two structures are different.
Here the solution sets of (1) and (2) are same. But the order in which they
occur are different.
4ti
For example Put n = 1 in (1), we get, = ~t"
4n
Put n = in (2), we get, 9 = ~r~
It is a convention to take that value of 9 whose absolute value is least as a
(principal value) to define the general solution.
Example 6.43: Solve : 2cos 9 + 3sin9 =
Solution:
2cos 2 9 + 3sin9 = => 2 (1  sin 2 9) + 3 sin9 =
=> 2sin 2 9  3sin9  2 =
=> (2 sin9 + 1) (sin 9  2) =
=> sin 9 = ~y~ (v sin9 = 2 is not possible)
=> sin 9 =  sin r
=^> sin 9 = sin I  t
^ e = n7t + (lf.f) ; weZ
Example 6.44: Solve : 2tan9  cot9 =  1
Solution:
2 tan9  cot9 =  1
2 tan 9  — "r =  1
tan9
^•2tan 2 9 + tan9l =0
(2 tan9  l)(tan9 + 1) =
220
2 tanG 1=0 or tanO +1=0
tan9 = 2 or tan 9 =  1
n
When tan9 =  1 =  tan t
tanG = tan   t
=>Q = nn + \~7
= nnT ; n e Z
When tan 9 = ~ = tan(3 (say)
.. 9 = nn + P
lf r
= nn + tan I ~
Hence 9 = nn  t or 9 = rat + tan" [ ~ J ', n e Z
Example 6.45: Solve : sin2x + sin6x + sin4x =
Solution:
sin2x + sin6x + sin4x = or (sin6x + sin2x) + sin4x =
or 2sin4x . cos2x + sin4x =
sin4x (2 cos2x + 1) =
nn
when sin4x = => 4x=nn or x = ~r ; we Z
_ _ — 1 n ( n] In
When 2 cos2x + 1 = => cos 2x = ~y =  cos t = cos I n — w I = cos ~r~
27r n
.". 2x = 2nn ± ~T" or x = n n ± t
Hence x = ~r~ or x = nn±^ ; n e Z
Example 6.46: Solve : 2sin jc + sin 2x = 2
2 2
Solution: 2 sin x + sin 2x = 2
2 2
.". sin 2x = 2 2sin *
2
= 2(1  sin x)
2 2
sin 2x = 2 cos x
221
2 2 2
4sin x cos x  2 cos x =
2 2 2
2(1  cos x) cos x  cos x =
2cos x  cos x =
2 2
cos x (2 cos x  1) =
cos x =
2 2^
cos x = cos 2
=> x = nn +^ , n & Z
COS X = 2 =
'%f
2 2 n
COS X = COS T
x = mn ± T , m e Z
Example 6.47: Solve : tan 2 9 + (l  a/3) tan9  a/3 =
=^> tan 2 9 + tan9  a/3 tanG a/3 =0
=> tanG (tan9 + 1)  a/3 (tanG + 1) =
=} (tanG +1) (tanG a/3 ) =
=^> tanG =  1
tanG = tan  j
G = nn ■
neZ
tanG = a/3
tanG = tan t
mn + T , m e Z
6.4.3 Solving equation of the form a cosG + £> sinG = c. where c <a +b
acosG + bsinG = c ■••(!)
Divide each term by ^a + b ,
la 2 + b 2
cosG +
V
Choose cosa =
sinG =
V« 2 + b 2 V
a 2 + b 2
sina = '
^\ja +b \ja +b
:. (1) becomes cosG cosa + sinG sina = cos(3
=> cos (G  a) = cos(3
=^> G  a = 2nn ± (3
=> G = 2nn + a + (3 , n e Z
Example 6.48: Solve : a/3 sin x + cosx = 2
and cos(3 =
V
2 , i2
222
2 2 2
Solution: This is of the form a cos* + b sinx = c, where c < a + b
So dividing the equation by a/(a/3~) +1 or 2
„ T a/3 . 1 , . 71 TC
We get ^ sin x + j coax =1 => sin t . sinx + cos t . cos x = 1
i.e. cosIjc — Tl = 1
cos i x  t I = cos
x — T = 2mtc +
i.e. x = 2wrc + T , « e Z
EXERCISE 6.7
(1) Find the principal value of the following equations:
1
(i) sin9 =
V2
(ii) 2 cos9 1=0 (iii) a/3 cot 6 = 1
1
i/1
' 2
(iv) a/3 sec9 = 2 (v) sinx =
(vii) sec x = 2
(2) Find the general solution of the following equation:
(vi) tan9 =
1
(i) sin29 = ^
(ii) tan9 =  a/3 (iii) cos 39 =
V3
V2
(3) Solve the following :
(i) sin3x = sinx (ii) sin Ax + sin2x = (iii) tan2x = tanx
(4) Solve the following:
? 1
(i) sin 2 9  2cos9 + 4=0
(iii) cosx + cos2x + cos3x =
(5) Solve the following:
(i) sin9 + cos9 = a/2
(iii) ^2 sec9 + tan9 = 1
2 2
(ii) cos x + sin x + cosx =
(iv) sin2x + sin4x = 2sin3x
(ii) sin9  cos9 =  a/2
(iv) cosec9  cot9 = a/3
223
6.5 Properties of Triangles
Consider a triangle ABC.
It has three angles A, B and C.
The sides opposite to the angles A, B, C are denoted
by the corresponding small letters a, b, c respectively.
Thus a = BC, b = CA, c = AB.
Fig. 6.21
We can establish number of formulae connecting these three angles and sides.
I. Sine formula:
In any triangle ABC, ^^
sin B ~~ sinC
of the circum circle of the triangle ABC.
In fig(6.22) O is the circumcentre of the triangle
ABC. R is the radius of the circumcircle. Draw OD
a
perpendicular to BC. Now BC = a, BD = ~
Clearly ABOC is an isosceles triangle.
We know that BOC = 2 BAC = 2A
.. I BOD =A
= 2R. Where R is the radius
Fig. 6.22
From the right angled triangle BOD,
Similarly, we can prove
II. Napier's formulae
In any triangle ABC
(1) tan
(2) tan
sinA =
BD all
= R _ R _
a
2R
2R sinA =
a
 a or sinA =
2R
b
sinB
= sinC = 2R
a
b c
sinA
(3)
tan"
AB
2
BC
2
CA
sinB
a — b
a + b
b  c
b + c
sinC
= 2R
C
cot
cot 2
c
a B
— ; — cot 7T are true
c + a 1
These are called Napier's formulae
224
Result (1):
tan 
AB ab
a + 1
C
coty
Proof: From sine formulae
ab
a + b
C
coty =
2R sinA  2R sinB
C
2R sinA + 2RsinB cot 2
.•. tan
=
sinA  sinB C
sinA + sinB cot 2
=
A+B . AB
2 COS r. Sill r. _
„ . A + B AB uuL 2
2 sin j cos ~
=
fA + B\ AB C
cot 1 2 J tan 2 cot 2
=
f™ °l A " B c
cot 1 90  2 1 tan 2 cot 9
=
C AB C AB
tan 2 tan ^ cot 2 = tan ^
A
B
ab C
cot
2  a + b " Ul 2
Similarly, we can prove other two results (2) and (3)
III. Cosine formulae
In any triangle ABC, the following results are true with usual notation
Results:
(1) a 2 = b 2 + c 2  2bc cosA
(2) b 2 = c 2 + a 2  lea cosB
(3) c 2 = a 2 + b 2  lab cosC
These are called cosine formulae
Result (1): a 2 = b 2 + c 2  2bc cos A
Proof:
Draw CD perpendicular to AB.
Now a 2 = BC 2 = CD 2 + BD 2
= (AC 2  AD 2 ) + (AB  AD) 2
= AC 2  AD 2 + AB 2 + AD 2  2AB x AD
= AC 2 + AB 2  2AB x (AC cosA)
2 2 2
a = b + c  2bc cosA
Fig. 6.23
225
Similarly we can prove the other results (2) and (3)
We can rewrite the formulae in different formats.
2 2 2 ,2
,2 2
? + c
cosA =
cosB = '
2 2
c + a ■
lea
cosC =
2,2 2
a + b  c
lab
IFF
IV. Projection formulae
In any triangle ABC
(1) a = b cos C + c cosB (2)b = c cosA + a cosC (3) c = a cosB + b cosA
are true with usual notations and these are called projection formulae.
Result (1): a = b cosC + c cosB
Proof:
In triangle ABC, draw AD perpendicular to BC.
From the right angled triangles ABD and ADC,
BD
cosB =
cosC =
But
AB
DC
AC
BD = AB x cosB
DC = AC x cosC
Fig. 6.24
BC = BD + DC = AB cosB + AC cosC
a = c cosB + b cosC
or a = b cosC + c cosB
Similarly, we can prove the other formulae (2) and (3)
V. Submultiple (half) angle formulae
In any triangle ABC, the following results are true.
„. . A . j(sb)(sc)
(1) sin 2 =\j bc
(2)sin =\
(4) cos y = ^
(6) cos y = '
(8) tan 2" ="
/(j c) (s  a)
J ca
,,, . C . I(sa)(sb)
(3) sin 2=^/ ab
s(s  a)
B Is(sb)
(5)cos 2 =^/ \ a
/j(j  c)
,,, A . j(sb)(sc)
(7) tan 9 = \ / x
2 \j s(s a)
, m , C . j(sa)(sb)
(9) tan 9 = A /
2 \j s(s c)
a + b + c
where j  ,,
/(i 1 — c) (5 — a)
V s(j  b)
The above results are called submultiple angles (or half angle) formulae.
226
Result (1): siny =
V 1 ^
c)
Proof :
We know that cos2A = 1 
 2sin 2 A
2sin 2 A = 1 
 cos2A
Replacing A by y ,
2 sin "2=1
cosA
= 1
u2 , 2
b + c 
2bc
2
a
a 2
(bc) 2
2bc  b  c + a
2bc
(a + b  c) (a  b + c)
2bc ~ 2bc
(a + b + c 2c)(a + b + c  2b)
2bc
(2s  2c) (2s  2b)
2bc
. . 2 A 2(sc)2(sb)
2sin 2 = 2^
2 A (sb)(sc)
sin 2 = fc
a + b + c = 2s
sin y = ±M ^
Since y is acute, sin y is always positive.
, . . (sb)(sc)
rhus sin 2=V te
Similarly we can prove the other two sine related formulae (2) and (3)
Result (4): cos y = M ^
Proof : We know that cos2A = 2cos A  1
2cos 2 A = 1 + cos2A
A T A
Replacing A by y , 2 cos y = 1 + cosA
b + c  a 2bc + b + c 
2bc ~ 2bc
(b + c) 2 a 2 (b + c + a)(b + c 
2
 a
 a)
2bc ~ 2bc
227
2 cos
cos
(b + c + a) (b + c + a  2a) 2s(2s  2a)
2bc ~ 2bc
2s x 2{s — a)
Result (7):
Proof:
cosy =
tany =
tany =
2bc
s(s — a)
be
js(sa)
o cosine related formulae (
. ksb)(s
c)
\j s(s  a)
A l(s
 i) (s  c)
sin 2 \J
£>c
cosy ^
/ (jfr)(jc)
j(j  a)
Similarly we can prove other two tangent related formulae (8) and (9)
VI. Area formulae (A denotes area of a triangle)
In any triangle ABC
(l)A = 2<a*sinC (2) A = ~ be sinA (3) A = ~ ca sinB
(4)A = ^ (5) A = 2R 2 sinA sinB sinC (6) A = a/^  a) (s  b) (s  c)
are true with the usual notations and these are called Area formulae.
Result (1): A = ~ ab sinC
Proof :
Draw AD perpendicular to BC
A = Area of triangle ABC
1
1
= 2 x BC x AD = 2 x BC x AC x sinC
1 AD
= 2®b sinC [v sinC = ~ttt => AD = AC x sinC]
Similarly we can prove the results (2) and (3)
Fig. 6.25
228
abc
Result (4): A = ^
Proof:
We know that
A = 2 ab sinC
1 u c
 2 cid 2R
abc
= 4R
C =2R
sinC
Result (5): A = 2R Z sin A sinB sinC
Proof:
We know that A = ~~ ab sinC
= ~ 2R sinA 2R sinB sinC ~:a = 2R sinA
= 2R 2 sinA sinB sinC b = 2R sinB
Result (6) Prove that A = ^js(s  a) (s  b) (s  c)
Proof:
We know that A = x ab sinC
1 C C
= y ab 2sin y cos y
= ^i'Ci'  a) (s  b) (s  c)
Example 6.49: In a triangle ABC prove that a sinA  b sinB = c sin(A  B)
Solution:
By sine formulae we have
a b c
sinA = sinB = sinC =
.". a = 2R sinA, b = 2R sinB, c = 2R sin C
a sinA  & sinB = 2RsinA sinA  2R sinB sinB
= 2R (sin 2 A  sin 2 B)
= 2R sin(A + B) sin(A  B)
= 2Rsin(180C)sin(AB)
229
= 2R sinC sin(A  B)
= c sin (A  B)
, ,«, , sin(A  B) a 2 b 2
Example 6.50: Prove that • , » r,*. = 5 —
Solution:
By sine formula ^ = = = ^ = 2R
a 2 fr 2 (2R sinA) 2  (2R sinB) 2
c 1 ~ (2R sinC) 2
4R 2 sin 2 A  4R 2 sin 2 B sin 2 A  sin 2 B
4R 2 sin 2 C
sin 2 C
sin(A + B) sin(A 
B)
[sinC = sii
sin 2 C
sin(A + B) sin(A 
B)
sin(A  B)
sin 2 (A +B)
~ sin(A + B)
Example 6.51: Prove that X a sin (B  C) =
Solution:
S a sin (BC) = a sin(B C) + b sin(C  A) + c sin (A  B)
= 2R sinA sin(BC) + 2R sinB sin(C  A) + 2R sinC sin(A  B)
sinA = sin(B + C), sinB = sin(C + A) ; sinC = sin(A + B)
= 2R sin(B + C) sin(B  C) + 2R sin(C + A) sin(C  A)
+ 2R sin(A +B) sin(A  B)
= 2R [sin 2 B  sin 2 C + sin 2 C  sin 2 A + sin 2 A  sin 2 B]
=
example 0.:
•>£: riove mat cos j = sin j
Solution:
b + c . A 2R sinB + 2R sinC . A
sin r, = <>r> • a sin
a 2 2R sinA 2
sinB + sinC A
= sinA sin 2
„ . B + C BC
2 sin 2 cus 1 a
A A sm 2
2 sin y cos y
230
. B+C BC
sin ~ cos ^
A
cosy
. fl80 fC\ B
sin 1 2 1 cus
C
2
A
cosy
• (— A^ B
C
sin 1 90 — j 1 cos j
A
cosy
BC
= cos 2
.• sin (90
f)
A
= cosy
Example
6.53: In
any triangle ABC prove that
a 2
sin (B 
sinA
 C) b 2 sin(C  A) c
x sinB x
2 sin(A 
sinC
B)
Solution:
2
a sin (B
Q
(2R sinA) 2 sin(B  C)
4R 2 sin ;
2 A sin
(B
C)
sinA
—
sinA
sinA
= 4R 2 sinA sin(B  C) = 4R 2 sin(B + C) sin(B  C)
= 4R 2 (sin 2 B  sin 2 C) = 4R 2 sin 2 B  4R 2 sin 2 C
= b 2 c 2
2 sin(C  A) 2 2
Similarly ^g = c  a
c 2 sin(A  B) 2 2
sh^ =fl "^
a 2 sin (B  C) b 2 sin(C  A) c 2 sin(A  B)
sinA ' sinB ' sinC
1? 2,2 2,2,2
= bc+ca+ab
=
EXERCISE 6.8
In any triangle ABC prove that
9 9 9 A 9 9 A
(1) a = (b + c) sin y + (b  c) cos y
231
(2) X a(b 2 + c 2 ) cosA = 3abc
(3) S a(sinB  sinC) =
(4) X (b + c) cosA = a + b + c
(5) a 3 sin(B  C) + b 3 sin(C  A) + c 3 sin(A  B) =
(6) a(fc cosC  c cosB) = b 2  c 2
2 2 2
cosA cos B cos C a + b + c
2abc
a
tanA
+ b +
2 , 2
c + a 
£ 2
tanB
~ j2 , 2
+ c 
2
 a
(7)
(8)
(9) If a cosA = b cosB then show that the triangle is either an isosceles
triangle or right angled triangle?
6.6. Solution of triangles
We know that a triangle has six parts (or elements). Consider a triangle
ABC. With usual symbols, the sides a, b, c and the angles A, B, C are parts of
the triangle ABC.
The process of finding the unknown parts of a
triangle is called the solution of triangle. If three parts of
a triangle (atleast one of which is a side) are given then
the other parts can be found. Here, we shall discuss the
following three types.
1) Any three sides (SSS) are given.
Fig. 6.26
2) Any one side and two angles (SAA) are given.
3) Any two sides and the included angle (SAS) are given.
Type I: Given three sides (SSS)
To solve this type, we can use any one of the following formulae.
(a) Cosine formula (b) Sine formula (c) Half angle formula .
It is better to use cosine formula if the sides are small, while use half angle
formula if the sides are large.
Example 6.54: Given a = 8, b = 9 s c = 10, find all the angles.
Solution: To find A, use the formula
2
a =
2 2
b + c  2bc cosA
cos A =
b 2 + c 2 a 2 81 + 100
2bc ~ 180
64
117
_ 180
A =
49° 28'
232
Similarly
cosB =
c 2 + a 2 b 2 100 + 6481
160
83
160
2ca
B= 58° 51'
A + B + C= 180°
.. C = 180° (49° 28' + 58° 51')
= 71° 41'
A = 49° 28', B = 58° 51', C = 71° 41'
Note: In the above example the numbers are smaller and hence we used cosine
formula.
Example 6.55: Given a = 31, b = 42, c = 57, find all the angles.
Solution: Since the sides are larger quantities, use half angle formulae
a + b + c
But
Thus
s =
= 65
log
tan"
A
A'
tan y =
1
(sb)(sc) = ^23x8^ 2
s(s  a) 1,65 x 34j
2 [log23 + log8  log65  log34]
2 [13617 + 0. 9031  1.8129  1.5315]
\ [1.0796] = \ [2 + 0.9204]
= \ I 2 + 0.9204J = 1 ,
4602
Y = 16°6'^A = 32° 12'
B
tan j =
(s  c) (s  a)
s(s  b)
1
65 x 23 J
lOE
B'
tan 2
= 2 [ lo g 8 + lo S 34 " lo S 65 " lo S 23 ]
= 2" [0.7400] = 2" [2+1.2600]
2 + 1.2600.
= 1 .6300
233
=> 2 = 23 ° 6 ' => B = 46 ° 12 '
C= 180(A + B) = 101°36'
Thus A= 32° 12' B = 46° 12' C=101°36'
Type II: Given one side and any two angles (SAA)
To solve this type, draw a sketch of the triangle roughly, for better
understanding and use sine formula.
Example 6.56: In a triangle ABC, A = 35° 17', C = 45° 13', b = 42.1. Solve the
triangle.
Solution:
The unknown parts are B, a, c
B = 180  (A + C) = 180  (35° 17' + 45° 13')
= 99° 30'
To find the sides, use sine formula
sinA
sinB sinC
6sinA 42.1 xsin35° 17'
a =
Again
sinB s in99° 30'
log a = log 42.1 + log sin35° 17'  log sin99° 30'
= 1.6243+ 1 .7616  1 .9940
= 1.3859 1.9940
= 1.3859 [1 + 0.9940] = 1.3919
a = 24.65
frsinC _ 42.1 xsin45° 13'
c ~ sinB ~ sin99'30'
logc = log 42.1 +logsin45° 13' log sin99° 30'
= 1.6243+ 1 .8511  1 .9940
Thus
= 1.4754 1 .9940
= 1.4754 [1+0.9940] = 1.4814
c = 30.3
B = 99° 30', a = 24.65, c = 30.3
234
Type III: Given two sides and the included angle (SAS)
Since two sides and the included angle are given, the third side can be
found by using the proper cosine formula. Then one can apply the sine formula
to calculate the other elements.
Example 6.57: Solve the triangle ABC if a = 5, b = 4 and C = 68°.
2
i
2
2 2 2
Solution: To find c, use c = a + b  lab cosC
= 25 + 16  2 x 5 x 4 cos 68°
= 4140x0.3746 = 26.016
c = 5.1
To find the other two angles, use sine formula.
b sinC 4 x sin68°
=> sinB = ^^=—^ l
log sinB = log 4 + log sin68° log 5.1
= 0.6021 + 1 .9672  .7075
= 0.56930.7075 = 0.1382
= 1 .8618
B = 46° 40'
=> A = 180 (B + C)=180 (114° 40')
= 65° 20'
Thus B = 46° 40', A = 65° 20', c = 5.1
Note: To find the angles A and B one can also use the tangent formula
AB ab C
tan — ~ — = — ~r cot^r
2 a + b 2
6.7 Inverse Trigonometrical functions (Inverse circular functions)
The quantities sin  x, cos  x, tan  x, ... are called inverse circular
functions, sin  x is an angle 9, whose sine is x. Similarly cos  x denotes an
angle whose cosine is x and so on. The principal value of an inverse function is
that value of the general value which is numerically the least. It may be positive
or negative. When there are two values, one is positive and the other is negative
such that they are numerically equal, then the principal value is the positive one.
_i fl\ n n
For example the principal values of cos I y I is t and not  t though
3
n\ 1
COS — a I  9
235
Note : sin  x is different from (sinx) ~ . sin  in sin  x denotes the inverse of
the circular function. But (sinx)  is the reciprocal of sinx i.e. ^— .
The Domain and Range of Inverse Trigonometrical functions are
given below:
Function
Domain
Range (Principal Value)
1.
y = sin  x
1<JC<1
n n
2.
y = cos x
1<X<1
0<y<n
3.
y = tan  x
R
% n
2<y<2
4.
y = cosec  x
x > 1 or i <  1
 2 ^ y ^ 2 ' y * °
5.
y = sec  x
x > 1 or x <  1
n
0<y<n; y^2
6.
y = cot x
R
0<y<n
Table 6.6
Example 6.58: Find the principal values of:
(i)Sin 1 (£) (ii)sec 1 ^ (hi) tan 1 (4
(v) cos   9 (vi) cosec  ( 2)
(iv) sin 1 ( 1)
Solution:
(i)
Then
• l f^\ ~ n n
Let sin lyl = y* where ~j <y< ~
sm
Ml
1
= y => sin y = 2 =
sin"
y=e
:. The principal value of sin I ~ J is ~z
(ii)
Let sec —j= = y, where < y < w , then,
sec ~ l H) = y^»~y = j 3
:. The principal value of sec
2 n n
= y => sec y = 77^ = sec g => v = g
_1 U
IS'
236
1 ( 1 ^ , n n
Let tan  —j= = y, where  — < y < —
(iii) lcl ian i  r
Then tan 1 ( 77=) = y => tarry =  ^ = tan ( J ^y = ^
:. The principal values of tan"  ~7f is t~~
(iv) Let sin (1) = y, where —~~ < x < ~~
Then, sin" (1) = y => siny =  1
1 = sin [ f J =>y = f
1 t
.". The principal value of sin ( 1) is  ~~
(v) Let cos  \ — j) = y, where < y < n, then
if f) 1
cos I — 2 J = .y=>cos;y = 2
cos;y = cost => cos y = cos I n — ~l =>y = l ~T
.". The principal value of cos  I  — I is
2; ia 3
1 "•
(vi) Let cosec ( 2) = y, where  ~~ < v <
_l , „. „ f — 7t l  7t
cosec (2)= y => cosec j = 2 = cosec I "T  ! ^>}) = 7~
1 — ^
.". The principal value of cosec (2) is —7
Example 6.59:
(i) If cot  ( — 1 =9, find the value of cos9 (ii) If sin  ( —J = tan  >.
find the value of x
Solution:
(i) cot 1 ( 7 J = => cot9 =  /. tanG = 7
7; _ u "" ^ LU _ 7
sec 9 = Vl+tan 2 9 = V*+49
237
sec 9 = 5^/2
=> cos9 = ^
l • l f^\ n 1 n
(11) tan x = sin lyl = Z •'• tan x = t
n 1 1
Properties of principal inverse Trigonometric functions:
Property (1):
(i) sin  (sinx) = x (ii) cos  (cosx) = x (iii) tan  (tanx) = x
(iv) cot  (cotx) = x (v) sec  (sec x)=x (vi) cosec  (cosec x)=x
Proof:
(i) Let sinx = y, then x = sin  (y) ... (1)
.". x = sin  (sinx) by (1)
Similarly, the other results may be proved.
Property (2):
(i) sin  I —J = cosec  x (ii) cos  I —J = sec  x
(iii) tan  I —J = cot  x (iv) cosec  I —J = sin  x
(v) sec  I —J = cos  x (vi) cot  (~) = tan  x
x) ■ ' \x
Proof:
(i) Let sin 1 f A = 9 => sin9 = 
^> cosec9 = x
^> 9 = cosec  (x)
^> sin I — I = cosec x
Similarly the other results can be proved.
Property (3):
(i) sin  ( x) =  sin  x (ii) cos  ( x) = n  cos  x
(iii) tan  ( x) =  tan  x (iv) cosec  ( x) =  cosec  x
(v) sec  ( x) = n  sec  x (vi) cot  ( x) =  cot  x
238
Proof:
(i) Let sin  ( x) = 9 .'.  x = sin9
=> x =  sin9
x = sin( 9)
=>  9 = sin" x
=> 9 =  sin" x
=> sin  ( x) =  sin" x
(ii) Let cos" ( x) = 9 => x = cos9
=> x =  cos9 = cos (n  9)
=> n  9 = cos  x
=> 9 = n  cos x
=> cos  ( x) = n — cos  x
Similarly the other results may be proved.
Property (4):
... . i i n _i _i n _j _j 7i
(l) sin x + cos x = y (n) tan x + cot x = y (m) sec x + cosec x = y
Proof:
(i)
Let
sin  x = 9 =>
x = sin9 = cos 1 z— 9
=>
l n Q
cos x = y  9
=>
l n • l
cos x = 2  sin x
• 1 It
=> sin x + cos x = 2
Similarly (ii) and (iii) can be proved.
Property (5):
If xy < 1, then tan 1 x + tan 1 y = tan 1 h ■*]
Proof: Let tan  x = 9i and tan  y = Q2 then tan9i = x and tan92 = y
tan9i+tan92 x + y
^>tan(9i + 92) = ~, — 7 — q — 7 — 7T = ~,
1 z 1  tan9l . tan92 I  xy
239
9j + 62 = tan
1 ( x + y
1 — xy.
1 1 1 ( x + y
=> tan x + tan y = tan
Note: Similarly, tan  x tan  y = tan  I 1
VL + xy.
Property (6): sin Ix + sin 1 y = sin l \_x \jly +y\lx \
Proof: Let 9i = sin  x and 62 = sin  y then sin9i = x and sin92 = y
=> sin(9i + 92) = sin9i cos92 + cos9i sin92
= (sinGi "ul  sin 92 + "\yl  sin 9l sin92)
= UVTV + yVT^]
=> 9i + 9 2 = sin 1 [x\lly 2 + y\jlx 2 ]
sin x + sin y = sin \_x\ly +y
V^ 2 ]
Example 6.60:
Prove that (i) tan I y J+tan I yr J =
Solution:
11
1 _L
tan
1±
(ii)cos
11
_1 27
5 + tan 5 = tan yy
11.
(i) tan \~\ + tan * 1 77J = tan
7 + 13
il i ,
_i 4 4 3
(ii) Let cos 7=9 then cos9 = t .'. tan9 = 7
= tan M90J =tan 1 k
= tan
_1 4 _! 3
cos j = tan 7
_i4 _i3 _i3 _ x 3 _i
.". cos 5+tan j = tan 7 + tan j = tan
4 + 5
_l27
y\ {3\i ~ tan 11
Example 6.61: Show that tan + tan y + tan z = tan
A) v4
1 fx + y + zxyz
Solution: tan x + tan y + tan z =tan
x + y
1  xy.
+ tan z
1  yz  zx  xy.
240
= tan
= tan
x + y
I —xy
+ Z
(x + y)z
1 — xy
= tan
X
iyi
z
xyz
1
xy
1
XV
xz
yz
1 — xy
x + y + g  xyz
1  xy  yz  zx_
—1 —1 ^
Example 6.62: Solve tan 2x + tan 3x = t
_1_ _1_ IT _1
tan 2x + tan 3x = t => tan
2x + 3x
l6x 2
= tan" 1 (1)
5x
l6x z
= 1 => 1  6x 2 = 5x .. 6x 2 + 5x  1 =
1
i.e. (x + 1) (6x  1) = => x =  1 or t
The negative value of x is rejected since it makes R.H.S. negative. .". x = t
Example 6.63:
Evaluate : (i) sinl cos  (tJJ (ii) cos I tan" jl (iii) sin lycos" t
3 3
Solution: (i) Let cos" t = 9 . Then, cos 6 = t
:. sin cos
! j =sin9 = Vlcos 2 9 =\/l
_9_
25
4
5
(ii) Let tan l f jj =9 then, tan9 = t
.'. cos I tan" jl = cos9 =
sec 9 Vl+tan 2 9 5
_i 4 4
(iii) Let cos t = 9 ; then cos 9 = t
sin
1 _!4"
2 cos 5
Example 6.64: Evaluate : cos
Solution:
. 6
= sin"
1 /lcosB
1
VTo
: cos
" . l 3 . i 5 "
sin t + sin tt
x ■ l 3 . A 3 A 4
Let sin t = A .•. sinA = t => cosA = t
5 c i y
Let sin" ~p? = B .". sinB = ~p? => cosB = tt
241
". cos
" ■ l3 . 1 5"
sin t + sin "pr
= cos (A + B) = cosA cosB  sinA sinB
fl 12 3 5) 33
_ V5 • 13 5 • 13j _ 65
EXERCISE 6.9
(1)
Find the principal value of
(i) sin  y (ii) cos  ( xj (iii) cosec  ( 1)
(iv) sec 1 ( y{2) (v) tan 1 (a/3) (vi) cos 1 ( j=)
(2)
Prove that (i) 2tan ( I = tan . (ii) 2tan x = sin 2
(iii) tan 1 (jj  tan 1 (jj = 4
(3)
Evaluate:
(i) cos( sin  tt I (ii)cos
(4) Prove the following:
l
(i) tan
(iii) tan
1  cosx
1 + cosx.
3
sin
x
: 2
(iii) tan I cos ~pyj (iv)
1 sin
cos
—1 2 —1
(ii) cos (2x  1) = 2cos x
3x  X
1 3jc 2
= 3tan x (iv) sin \2x "\j 1  x ) = 2sin >
(5) Prove that 2tan [ ~ = tan ! f ?
12
(6) Prove that tan
(7) Solve : tan 1
'if
xl
x 2
tan
+ tan"
1 l m — n \ _n
rn + nj 4
\{x + 1 \ 71
_i / 2x \ _i f 1 — xf 1 1 IT
(8) Solve tan  : ,  + cot  ,,,. J = T , where x >
1x'
x + 2
2
2x
■4
K
3
1 1 1 4
(9) Solve : tan (x + 1) + tan (x  1) = tan y
(10) Prove the following:
(i) cos 1 x + cos~ l y = cos 1 [xj  "\/ 1  x 2 "\/ 1  )> 2 ]
(ii) sin 1 x  sin 1 ); = sin 1 \_x\jl y y\jlx \
(iii) cos 1 x  cos 1 ;y = cos 1 [xy + V 1 * 2 • V 1 ~y 2 J
242
OBJECTIVE TYPE QUESTIONS
Choose the correct or most suitable answer
(1) The order of matrix B = [1 2 5 7] is
(l)lx
4
(2) 4 x 1
(3) 2 x 1
(4) 1 x 1
(2)
Number of elements in a matrix of order 2 x 3 is
(1) 5 (2) 2
(3) 3 (4) 6
(3)
If A =
"214"
_3 2 1.
and X + A =
then matrix X is
(1)
"214
.3 2 1
(2)
"_ 2  1  4~
3 2 1.
(3)
~2 1 4
_ 3 2 1
(4)
"2 1 4"
_3 2 1.
" 7"
(4)
The product of the matrices [7 5
31
3
?
is equal to
(1) [70] (2) [49]
(3) [15] (4) 70
"V2
o"
(5)
The type of the matrix
^3
V3.
is
(1) a scalar matrix
(2) a diagonal matrix
(3) a unit matrix
rm
(4) diagonal and scalar
(6)
if [:
I y
c 1]
A'
.3.
= [i:
5] then the va
ueo
f a is
(1)5 (2)2 (3) + 3 (4) ±4
(7) Matrix A is of order 2x3 and B is of order 3x2 then order of matrix B A
is
(1)3x3 (2)2x3 (3)2x2 (4)3x2
(8) If [3 1 2]B = [5 6] the order of matrix Bis
(1) 3 x 1 (2) 1 x 3 (3) 3 x 2 (4) 1 x 1
(9) The true statements of the following are
(i) Every unit matrix is a scalar matrix but a scalar matrix need not be a
unit matrix,
(ii) Every scalar matrix is a diagonal matrix but a diagonal matrix need
not be a scalar matrix.
243
(iii) Every diagonal matrix is a square matrix but a square matrix need
not be a diagonal matrix.
(1) (i), (ii), (iii) (2) (i) and (ii) (3) (ii) and (iii) (4) (iii) and (i)
(10) The matrix
"8
5
r
6
4
_0
2_
IS
(1) the upper triangular
(3) square matrix
2 3
(11) The minor of 2 in
(DO
6
is
(2)1
23 5
6 4
1 5 7
(1)  18 (2)18
(12) The cof actor of  7 in
(2) lower triangular
(4) null matrix
is
(3)2
(13) IfA =
aj bi
c l
a 2 h
c 2
«3 h
c 3
(3) 7
and I A I = 2 then I 3 A I is
(4) 3
(4)7
(1)54 (2)6 (3)27 (4) 54
(14) In a third order determinant the cofactor of a 2 3 is equal to the minor of
a 2 3 then the value of the minor is
(4)0
(4) x =
(1)1
(15) The solution of
(2)i
2x 3
2 3
= 0is
(3) A
(1)JC=1
(2) ;
1 1
c = 2
1
(3) x = 3
(16) The value of
2x 2y
lz is
3x 3y
3z
(1)1
1
2 3
(2) >
cyz
3 1
2
(3) x + y + z
(17) IfA =
3
2
1 2
3 1
then
1 2
2 3
3
1
ie equal to
(1)A
(2)
A
(3)3A
(4)0
(4)  3A
244
(18) The value of the determinant
1
2
3
7
6
5
1
2
3
IS
(1) (2) 5 (3) 10
(19) If A is a square matrix of order 3 then I kA I is
(4)  10
(l)fclAI
(2)  k 1 A
(3) r i a i
(4)*r
IAI
1 4 3
2
8 6
(20) IfA =
11 5
3 21
and Aj =
2
6
2 10
4 2
then
(1) Aj = 2A (2) A l = 4A
(3) A l = 8A
(4)A =
8Aj
7 6 1
7 6
1
(21) IfA 1 =
5 3 8
and A 2 =
8 2
4
then
8 2 4
If
6
16
(l)Aj=2A 2 (2)A 2 = 2Aj (3) Aj = 2A 2 (4)A,=2A 2
(22) Two rows of a determinant A are identical when x =  a then the factor of
A is
(1) x + a (2) x — a
(23) The factor of the determinant
(3) (x + ay (4) (x  ay
x 6 1
2 3i i3 is
 3 2x x + 2
(1) x + 2 (2) x  3 (3) 2jc + 1 (4) jc + 3
(24) If all the three rows are identical in a determinant A on putting x = a then
the factor of A is
(l)xa (2)x + a
(25) The factor of the determinant
(l)jt (2)x + b
(26) The value of the determinant
(1) abc (2)
o)( X  a y
(4) (jc + a y
x + a
b c
a
x + b c
is
a
b x + c
(3)x + c
(4)xa + b + c
a
2
b
is
c
(3) a 2 b 2 c 2
(4)  abc
245
(27) The value of the product
(2) 56
1 2
3 1
(1)56
(28) IfA =
2
1 4
(3)l
is
a \
*1
C\
a 2
&2
c 2
«3
h
c 3
(4) 63
and Aj, Bj, Cj are the cofactors of aj, ft 1; c l
then a l A 2 + b l B 2 + c { C 2 is equal to
(1)A (2)0 (3) A (4) A 2
(29) Given that the value of a third order determinant is 1 1 then the value of
the determinant formed by the respective cofactors as its elements will
be
(1)11 (2)121 (3)1331 (4)0
(l+ax) 2 (l+ay) 2 (1 + azf
(2) 121
(30) A factor of the determinant
(l)x + y (2)a + b
(l+bx) 2 (l+by) 2 (1 + ftz) 2
(1 + ex) 2 (1 + cy) 2 (1 + cz) 2
is
(3)jcy
(4) a + b + c
»
»
^>
^>
>
»
(31) The position vector of A is 2 i + 3 j + 4 A; , AB = 5 i + 1 j + 6k
then position vector of B is
(1) 7 j + 10 v + 10 k
> > >
(2) 7 j 10 j + 10 it
(3)7?
+ 10 7 10 it (4)
7
j +10 7
10 A
>
(32)
>
If a is a non zero vector and & is a scalar such that
k a
= 1 then Ac is
equal to
/1 \
>
1
//I \ _i_ 
i
(1) a \L) i p;
>
a
(4)±
>
a
(33)
> > >■ >■
Let a , b be the vectors AB , BC determined by two adjacent sides
>
of regular hexagon ABCDEF. The vector represented by EF is
(1)
a 
ft (2) a + ft (3) :
I a
;4>
ft
246
(34) If AB =k AC where k is a scalar then
(1) A, B, C are collinear (2) A, B, C are coplanar
(3) AB , AC have the same magnitude (4) A, B, C are coincident
(35) The position vectors of A and B are a and b . P divides AB in the ratio
3 : 1 . Q is the mid point of AP. The position vector of Q is
> > » — » » > » — »
5a + 3fr 3 a + 5 fc 5 a + 3& 3 a + fc
(,U g (A) 2 ^ ' 4 ^ ' 4
(36) If G is the centriod of a triangle ABC and O is any other point then
OA + OB + OC is equal to
(1) O (2) OG (3) 3 OG (4) 4 OG
(37) If G is the centriod of a triangle ABC then GA + GB + GC is equal
to
/ \ * ~* +
(1) 3 \a + b + c) (2) OG (3) O (4) 3
(38) If G is the centriod of a triangle ABC and G' is the centroid of triangle
A' B' C ' then A A' + BB' + CC ' =
(1) GG' (2)3GG' (3)2GG' (4)4GG'
» »
(39) If the initial point of vector  2 i  3 j is ( 1, 5, 8) then the terminal
point is
> » > » > »
(1) 3 j + 2; + 8 * (2)  3 j + 2 y + 8 k
> > > » > >
(3)  3 i  2 y  8 * (4) 3 i + 2 y  8 k
(40) Which of the following vectors has the same direction as the vector
i 2y
(1) i +2y (2)2/ +4y (3)  3 » +67 4)3/ 67
247
(41)
If a = j + _/'  2 k , b =  i + 2 j + k ,
» » » »
2 fc , then a unit vector parallel to a + b + c is
c = j  2 y +
'"^^^r^ 1 ^^
(4,^3^
(42)
> — » — » — » » — » — » — »
If a =2i +y  8 it and fc = i + 3 j  4 it
» »
of a + b =
then the magnitude
(1)13 (2)13/3 (3)3/13 (4)4/13
(43) If the position vectors of P and Q are
> » > » » >
2i + 3 y 1 k , 4 i  3 j + 4 k , then the direction cosines of PQ
are
2 6 11 2 6 11
( } VT61 ' VT61 ' VT61 ( } Vi6i 'VT6T 'VT6T
(3)2,6,11 (4)1,2,3
ax 2 3
(44) If (x + 2)(2x3) = ^2 + 2^ thena =
(1)4 (2)5 (3)7 (4)8
(45) If nPr = 720 nCr, then the value of r is
(1)6 (2)5 (3)4 (4)7
(46) How many different arrangements can be made out of letters of words
ENGINEERING
11! 11! 11!
(Dll! (2) ^W (3) 3H2! < 4 >!T
(47) The number of 4 digit numbers, that can be formed by the digits
3, 4, 5, 6, 7, 8, and no digit is being repeated, is
(1)720 (2)840 (3)280 (4)560
(48) The number of diagonals that can be drawn by joining the vertices of an
octagon is
(1)28 (2)48 (3)20 (4)24
(49) A polygon has 44 diagonals then the number of its sides is
(1)11 (2)7 (3)8 (4)12
248
(50) 20 persons are invited for a party. The number of ways in which they and
the host can be seated at a circular table if two particular persons be
seated on either side of the host is equal to
(1)18! 2! (2) 18! 3! (3) 19! 2! (4) 20! 2!
(51) If n is a positive integer then the number of terms in the expansion of
(x + a) n is
(1) n (2) n  1 (3) n + 1 (4) n + 2
(52) The values of wCO  nC 1 + nC2  nC3 +...( 1)" . nCn is
(1)2" +1 (2)n (3)2" (4)0
(53) The sum of the coefficients in the expansion of (1  x) is
(1) (2) 1 (3) 10 2 (4) 1024
24
(54) The largest coefficient in the expansion of (1 + x) is
(1)24C24 (2)24C13 (3) 24C12 (4)24Cll
r 2n 1 8
(55) The total number of terms in the expansion of i(a + b) \ is
(1)11 (2)36 (3)37 (4)35
(56) Sum of the binomial coefficients is
(1) In (2) n (3) 2" (4) n + 17
8
(57) The last term in the expansion of (2 + sj3) is
(1)81 (2)27 (3)^/3 (4)3
(58) If a, b, c are in A.P., then 3", 3 b , 3 C axe in
(1) A.P. (2) G.P.
(3) H.P.
(4) A.P. and G.P.
(59)
If the « th term of an A.P. is (In  1) :
then the sum of
n terms is
(1) n 2  1 (2) (2m  1)
(3)n 2
(4) n 2 + 1
(60)
2
The sum of n terms of an A.P. is n .
Then its common difference is
(1)2 (2) 2
(3) + 2
(4)1
(61)
The sum to the first 25 terms of the :
series 1 + 2 + 3 . .
is
(1) 305 (2) 325
(3) 315
(4) 335
(62)
The n term of the series 3 + 7 + 13
+ 21+31 + ... .
is
(1) An  1 (2) n 2 + In
(3) (n 2 + n +
1) (4) (n 3 + 2)
(63)
What number must be added to 5,
13 and 29 so that sum may form a
G.P?
(1) 2 (2) 3
(3)4
(4)5
249
(64) The third term of a G.P. is 5, the product of its first five terms is
(1)25 (2)625 (3)3125 (4)625x25
(65) The first term of a G.P. is 1. The sum of third and fifth terms is 90. Find
the common ratio of the G.P.
(1)±2 (2)VlO (3) + 3 (4) 3
(66) When the terms of a G.P. are written in reverse order the progression
formed is
(l)A.P. (2) G.P. (3)H.P. (4) A.P. and H.P.
(67) If A, G, H are respectively arithmetic mean, geometric mean and
harmonic mean then
(1)A>G>H (2)A<G>H (3)A<G<H (4) A > G < H
(68) The A.M. between two numbers is 5 and the G.M. is 4. Then H.M.
between them is
(1)3 j (2)1 (3)9 (4)1 \
(69) If a, b, c are in A.P. as well as in G.P. then
(1) a = b * c (2) a * b = c (3) a * b * c (4) a = b = c
(70) The A.M., G.M. and H.M. between two positive numbers a and b are
equal then
(1) a = b (2) ab=\ (3)a>b (4)a<b
2 3
X X
(71) e x = 1 + x + j\ + 3T + is va lid for
(1)1<x<1 (2) 1 <jc< 1 (3) all real x (4)jc>0
(72) <? log * is equal to
(l)x (2)1 (3)e (4)log^
(73) The equation of xaxis is
(1)jc = (2)x = 0,;y = (3)y = (4)jc = 4
(74) The slope of the straight line 2x  3y + 1 = is
2 3 2 3
(1)— (2)— (3)3 (4) 2
(75) The y  intercept of the straight line 3x + 2y  1 = is
(1)2 (2)3 {3)\ (A)\
(76) Which of the following has the greatest jintercept in magnitude?
(l)2x + 3y = 4 (2)x + 2y = 3 (3)3x + 4y = 5 (A)Ax + 5y = 6
250
(77) If the equation of the straight line is y = \J3 x + 4, then the angle made by
the straight line with the positive direction of xaxis is
(1)45° (2)30° (3)60° (4)90°
(78) If the straight lines a^x + b^y + c^ = and a 2 x + b 2 y + c 2 = are
perpendicular, then
a
V^=~V 2 ^7T 2 =V 2 V)a,a 2 =  bl b 2 (4) =y 2 =
(79) Which of the following is a parallel line to 3x + Ay + 5 = 0?
(1) 4jc + 3;y + 6 = (2)3jc4j + 6 =
(3) 4x  3y + 9 = (4) 3x + Ay + 6 =
(80) Which of the following is the equation of a straight line that is neither
parallel nor perpendicular to the straight line given by x + y =
(l)y = x (2)yx + 2 = (3)2y = 4x+l (4)y+x + 2 =
(81) The equation of the straight line containing the point (2, 1) and parallel
to 4x  2y = 3 is
(l)y = 2x + 5 (2)y = 2xl (3)y = x2 (4)y = ^x
(82) Equation of two parallel straight lines differ by
(l)xterm (2) y term (3) constant term (4)xyterm
2
(83) If the slope of a straight line is t , then the slope of the line perpendicular
to it, is
2 2 3 3
(1) 3 (2) J (3) 2 (4) J
(84) The graph of xy = is
(1) a point (2) aline
(3) a pair of intersecting lines (4) a pair of parallel lines
2 2
(85) If the pair of straight lines given by ax + 2hxy + by = are
perpendicular then
(l)ab = (2)a + b = (3)ab = (4)a =
251
2
(86) When h =ab the angle between pair of straight lines
2 2
ax + 2hxy + by = is
(l)f (2)f (3)f (4)0°
2 2
(87) If 2x + 3yx  cy = represents a pair of perpendicular lines then c =
(l)2 (2)  (3)2 (4) 
2 2
(88) If 2x + kxy + Ay =0 represents a pair of parallel lines then k =
(1)±32 (2) ±2^2 (3) +4 a/2 (4) ± 8
2 2
(89) The condition for ax + 2hxy + by + 2gx + 2fy + c = to represent a pair
of straight lines is
(\)abc + 2fgh bf ag 2 ch 2 = (2) abc  2fgh  ag 2  bf  ch 2 =0
(3) abc + 2fgh  ah 2 bg 2 cf = (4) abc + 2fgh  af 2  bg 2  ch 2 =0
(90) The length of the diameter of a circle with centre (2, 1) and passing
through the point (2, 1) is
(1)4 (2)8 (3)4^/5 (4)2
2 2
(91) Given that (1,  1) is the centre of the circle x +y + ax + by  9 = 0. Its
radius is
(1)3 (2) >/2 (3) VII (4)11
(92) The equation of a circle with centre (0, 0) and passing through the point
(5,0) is
(l)x 2 + y 2 10x = (2)x 2 + y 2 = 25
(3) x 2 + y 2 + 10* = (4)x 2 + y 2 10y =
2 2
(93) The radius of the circle x +y 2x + 4j4 = 0is
(1)1 (2)2 (3)3 (4)4
2 2
(94) The centre of the circle x +y +2x4j4 = 0is
(1)(2,4) (2) (1,2) (3) (1,2) (4) (2, 4)
252
(95) If 2x + 3y = and 3x  2y = are the equations of two diameters of a
circle, then its centre is
(l)(l,2) (2) (2, 3) (3) (0,0) (4) (3, 2)
2 2
(96) If the line y = 2x  c is a tangent to the circle x + y =5, then the value
of c is
(1)±5 (2)+V5 (3) + 5a/5 (4)±5y/2
2 2
(97) The length of the tangent from (4, 5) to the circle x +y = 25 is
(1)5 (2)4 (3)25 (4)16
(98) If the circle has both x and y axes as tangents and has radius 1 unit then
the equation of the circle is
(l)x 2 + (yl) 2 =l (2)x 2 + y 2 =l
(3)(xl) 2 + (yl) 2 =l (4)(xl) 2 + /=l
2 2
(99) Which of the following point lies inside the circle x + y  4x+2y  5=0
(1)(5, 10) (2) (5, 7) (3) (9,0) (4) (1, 1)
(100) The number of tangents that can be drawn from a point to the circle is
(1)1 (2)2 (3)3 (4)4
(101) If two circles touch each other externally then the distance between their
centres is
(102) The number of points in which two circles touch each other internally is
(1)1 (2)2 (3)0 (4)3
(103) One radian is equal to (interms of degree)
180° ,„ n ,„ 180 _ 11
(Drr (2)— (3)— (4)
11 w 180° v; n ^'180°
(104) An angle between 0° and  90° has its terminal side in
(1)1 quadrant (2) III quadrant (3) IV quadrant (4) II quadrant
253
(105) tTq of a complete rotation clockwise is
(1)1° (2) 360° (3) 90° (4)1°
(106) If the terminal side is collinear with the initial side in the opposite
direction then the angle included is
(1) 0° (2) 90° (3) 180° (4) 270°
(107) Area of triangle ABC is
(1)2 ab cosC (2) 2 ab sinC (3) y ab cosC (4) y be sinB
(108) The product s(s  a) (s  b) (s  c) is equal to
A
(1)A (2) A 2 (3)2A (4)^
(109) In any triangle ABC, A is
abc abc abc
(I) abc (2) W (3)2R (4)r
(110) In triangle ABC, the value of sinA sinB sinC is
A ... A ... A A
— 2 ( 4 ) —
2R 2 4R
d>2R ( 2 >4R 0)^2 (4) AP 2
(111) cosB is equal to
c 1 + a 2  b 2 ^
(1) lea {V> 2bc (3) lab (4) lab
c 2 + a 2  b 2 ... c 2 + b 2  a 2 ... a 2 + b 2  c 2 . .. a 2 + b 2 + c 2
254
(1) (i)
■2 3 4'
3 4 5
4 5 6.
(ii)
ANSWERS
EXERCISE 1.1
1 2 3"
2 4 6
3 6 9.
(2) x = 0, j = 7, z = 3
(4) (i)
(v)
(6) X =
^ 3 5 1
27
12
4
4
4
6 10.
(ii)
_ 2 12.
(iii)
_6 4
"6 f
"6 1

"9 "
_5 2 _
(vi)
_5 2
(vii)
.14 16.
"22
1"
"2 5 1"
2 3
3
,Y =
11 3
_2 1
3_
_
7
2
(iv)
(viii)
4 4
6 4 .
8 '
18 7.
(8) k = 1
(10) jc=1,3 (ll)x = 2,5 (13) 
EXERCISE 1.2
9 3'
6 7.
(14)jc=l,y = 4
27
(1)0 (2) (i) nonsingular (ii) singular (3) (i)x = ~3~ (ii)x = 9
(4) (i)0 (ii) (6)a 3 + 3a 2
EXERCISE 1.3
(3) x = 0, 0,  (a + b + c) (4) (a  b) (b  c) (c  a) (ab + be + ca)
EXERCISE 2.1
> > > > > >
(1) AC = a + b , BD = b  a
EXERCISE 2.2
(1) 5~t +5~f +5~k , 5^/3
(2) VI 85
255
> > > > ► > > >
(3) AB =  3 i  j 5 k ; BC = 4 j  7 j + 7 jfe ;
>• > > >
CA = i +8 j 2*
AB = ^35 , BC = VTTi , CA = ^69
m± 17? A lot «»2(t + ? + t)
— >• > > > M 5 iO
(13) pq =4, 5; + n* ;^^,^J
(16) noncoplanar vectors.
EXERCISE 3.1
1 _ 1 20 _ _D_
(1) 2(jc1) 2(x+l) {2 'x3 x2
3 _ _7_ 13
( ' 2(x  1) "x2 + 2(jc  3)
( 4 ) 9fr7T) " 9(^T2) " ^7^2
_ 4 4 1
^ 9(* + 2) + 9(xl) 3 (xl) 2
2 3 2
(6) 25(x^2) + i^T^2 "25(xT3)
^7 J_ 9
(7) 2x + x 2 + 2(x + 2)
2 3 9
(8) 7T^ +
(x  2) ( x _ 2) 2 (jc  2) J
, Q , 1 4x8
^ 5(x + 2) + 5(x 2 +1)
256
(10) 1 (x ~ 3)
^ 1U )2(x + l) 2(x 2 +1)
x5 4
A' 
(11) ^l^T + 3.2
(12) 1777+^
EXERCISE 3.2
(1) 378 (2) 42 (3) 600 (4) 1320 (5) 42840
(6)512 (7)153 (8) (i) 27216 (ii) 90000 (9)5x5! (10)21
(11) 2 5 (12)9000 (13) (i) 125 (ii) 60 (14) 2 5
EXERCISE 3.3
(1) (i)60 (ii) 2730 (iii) 120 (iv)^f (v) 15120
(2) 23 (3)4
(4)41 (7)172800 (8)5040 (9)60 (10)93324 (11)34650
(12) (i) 840 (ii) 20 (13) 9000 (14) 4 5 (15) (i) 8! (ii) 7! (16) ^r
EXERCISE 3.4
(1) (i) 45 (ii) 4950 (iii) 1 (2) 23 (3) 3 (4) 45
(5) (i) 12 (ii) 8 (6) 19 (7) 7
EXERCISE 3.5
(1) 66 (2)200 (3)210 (4)425 (5) (i) 15 C n (ii) 14 C 10 (iii) i 4 C n
(6)780 (7) (i)40 (ii) 116 (8)1540 (9)817190
EXERCISE 3.7
(1) (i) 243a 5 + 2025 A + 6750a V + 1 1250a V + 9375a6 4 + 3125 b 5
(ii) a 5  10 A + 40a V  80a V + Wab 4  32b 5
(iii) 32jc 5  240jc 6 + 720jc 7  1080jc 8 + 810x 9  243.x
10
257
,. , 11 llx 10 55x 9 165x 8 330x 7 462x 6 462x 5 330x 4
(IV) x + — — + —j + —3 + —4 + — ^ + — g + ~y~
J j y y y y y
165x 3 55x 2 llx J_
+ / + / V° V 1
(v) x 12 + 12x 10 y 3 + 60x 8 / + 16Q*V + 240x 4 ;y 12 + 192*V 5 + 64j 18
,, n 4 2 , . 7/2 3/2 , , 3 3 , A 5/2 7/2 , 2 4
(vi) x>>+4x j + 6x j + 4x y + x y
(2) (i) 58 ^2 (ii) 152 (iii) 352
(iv) 128a 3 + 4320a 2 + 9720a + 1458 (v) 5822 ^3
(3) (i) 1030301 (ii) 970299 (4) 0.9940
4 , 2 !6C8 a 8
(5) (i) 8C4 2V 2 (ii) 16C 8 (iii) j —
x
(iv) 13C 6 .2 6 x 7 / and  13C 7 . 2 7 x 6 /
(v) 17C 8 .2 8 y and 17C 9 .2 9 yj
x x
(7)  165 (8) (i) 7920 (ii) 2268 (iii) i 2 C 4 [J 9 8
(9) r=3 (10) 7, 14
EXERCISE 4.1
(1) (i) 25,  125, 625,  3125 and 15625 (ii) § ,  , y , 21, y
(iii) 1,12, 23, 34, 45 (iv)  , f , , , f
..222 1 4 1 16 25
(v)j , 0, 3 , 0, 3 (vi) 3 , 9 , 3 , gj, 243
(2) (i)y , y (ii) 1,0 (iii)y , yf (iv) 64, 512
5 17 37
(3) 0, y 8, y, 24, y
258
(4) (i) 2, 2, 1, 0, 1
(iii) 1,2,6,24, 120
(ii) 1, 2, 3, 5, 8
(iv) 1,1,5, 13,41
(5)
1
1
3"
(6)4 [5 n l]
EXERCISE 4.2
(1) (i) 4, 7, 10, 13, 16 (ii) 5, 7, 9, 11, 13, 15
(7)doo(2 10 °D
(4)
19
(2) (i) 10 (ii) 1 (m)P
(6) 6, 24 (10) 2, 3, 6 (or) 6, 3, 2
EXERCISE 4.3
CD ©^
x
; + ■
40 160
"+...
(ii)
1
>/6
2 3
£ X_ IX
1+ 6 + 18 + 324 +
H9.7.5 12
(2) (i) 10.01 (ii) 0.2 (5)^
... (r+1) (r+2) (r+3) r
® L23 x
EXERCISE 5.1
(1) x 2 + /2jc + 8)'19 = (2)3x + j = 2
(3) (i)f=l (ii)P(l,2) (4)/24x 2 =
(7) (i) x 2 + y 2 + x  3y + 2 = (ii) 15x 2 + I5y 2 + 66x  96y + 207 =
EXERCISE 5.2
(1) 4jc7y10 = (2)y = 3x + 4 (3)xy = 6
(4) llxy = 27 (5)2x + y = 6;x + 2y = 6 (6)x + 3y = 8
(7) 3x  2y = ; 2x  y = and 5x  3y =
14
(8)jr^ units
VT3
(9) 2jc3)'+12 = (10)9jc8)'+10 = 0;2jcj =
6
(1 1) 2x  3y = 6; 3x  2y = 6 (12) x intercept ~ ; y intercept 2
(13) (8, 0) and ( 2, 0) (14) 3^2 units
259
EXERCISE 5.3
(1) f (3)3;c + 2y+l=0 (4)jcyl=0 (5) (1, 2)
(6) k = 9 (7) 2 mits (8)p=l;p = 2 (9) 28x + ly  74 =
(10) 5x + 3y + 8 = (ll)x + y=l (12) 5x + 3y + 5 = (14)J;
(17) a = 5 (18)a = Tf d9) (2, f) (21) d, 12) (22) ( 4,  3)
EXERCISE 5.4
(1) a = 2 ; c =  3 (2) nl3 (4) 1 (6) 2jc 2  3xy 2y 2 =
(7) 3x 2 + 7xy + 2/  4x + ly  15 =
(8) k =  1 ; Ax  3y + 1=0 and 3x + 4y 1=0 ; nil
(9) C = 2 ; 6jc  2y + 1 = and 2x y + 2 = ; tan" 1 (1/7)
(10) £ = 10 ; 3jc2j+l=0and4x + 5)' + 3 =
EXERCISE 5.5
(1) (i) (0, 0) ; 1 (ii) (2, 3) ; a/22 (iii) (4, 3) ; 7
f 2 T\ 2V5 ,—
(iv) r3'3"J ; 3 (v)(4,4);VT0
(2) a = 4 ; b = 2 ; 2x 2 + 2y 2 + Ax + Ay  1 =
(3) x 2 + y 2 Ax6y+U=0
(A) x 2 + y 2 6x Ay 12 =
(5) jc 2 + /14jc + 6)' + 42 =
(6) x 2 + y 2 + 8xl0y + 25 =
(7) 2\jlb n unit ; 10^ square units
(8) x 2 + y 2  Y2x + 11 = ; x 2 + y 2 + Ax  21 =
(9) x 2 + y 2 3x6y+l0 =
260
(10) x 2 + y 2 =l
(11) x 2 + y 2 5xy + 4 =
(12) x 2 + y 2 6x8y+l5 =
(13) x 2 + y 2 4x2y5 =
(14) 16x 2 + I6y 2 = 1
3 3
(15) x = 2 cos9;>' = 2' sin 9
EXERCISE 5.6
(1) 2^5 units (3) y  1 = (4) outside
(5) (0, 0) and (4,  3) lies inside; ( 2, 1) lies outside
(6) (0, 2) ; (2, 0) (7) 2x + y = ± 3^5 (8) 5a/2 units
(9) (i)x 2 + ;y 2 10jt:12;y + 25 = (ii) x 2 + y 2  10*  12y + 36 =
(10) 4x + 3y + 6 =
(12) x5y+19 =
(ll)(i) x + y = ±4>/2 (u)xy = ±4^[2
(13) + 40
EXERCISE 5.7
, 1 5
(14)144
(3) J + flx 6^39 =
(4) x 2 + y 2 8x +12^49 =
(6) (i) x 2 + y 2  2x + 2y + 1 = (ii) x 2 + y 2  6x  4y  44 =
(7) x 2 + y 2  \6x  18y  4 =
(8) 3x 2 + 3j 2  14jc + 23y  15 =
EXERCISE 6.1
(1) (i)t (n)"5 (m) 5— (iv)— 5— (v)— ^7— (vi)
g W 9 v"V 9 Viv; 9 vv; jg
(2) (i) 22° 30' (ii) 648° (iii)  171°48' (app.) (iv) 105°
(3) (i) Qj (ii) Q 3 (iii) Qi
EXERCISE 6.2
1331
24
(1)
276
(2) (i)  sin 60° (ii)  cos40° (iii) tanl0° (iv)  tan60°
(v)cosec60° (vi)sin30° (vii) cos 30°
261
(5) (i)  cosec A (ii)  sec A (iii)  cotA (iv) cosA (v) tan A
1
(8) (i)
V2
(v)l (vi)
a/3
.... \/3
(")" 2
(vii) 1
(iii)  a/2
1
(iv) 2
(viii)
V3
(10) (i)y (ii) 1 (iii)0 (iv r 2 4 ^ 6
(v)5 (vi)2 (vii)
25
(viii) 1 1 (ix) J2
(1) (i)^
(ii)
w 120
EXERCISE 6.4
\/6a/2
(iii) 2 + V3 (iv)^^
, A a/2+1 a/6 + A /2 x/2^/3 a/6 + A /2
(8) vv 2
(12) a/2 + 1
(3) i
(14)^^
(6) (i) ^
(1) (i) sin69 + sin29
(iv) cos2A  cos4 A
1
EXERCISE 6.5
^125
EXERCISE 6.6
(ii) cos 149 + cos29
(v) sin9A  sin3A
1
(iii) sinlOO  sin49
(vi>2 [sinl39 + sin59]
1
(vii) ^ [sin2A  sin A] (viii) y [sin6A + sin A] (ix) y [cos39+cos 9/3]
(2) (i) 2sin9A cos4A (ii) 2 cos9A sin4A (iii) 2cos9A cos4A
(v) 2 cos42° sin 10°
(viii) 2 sin30° cos20 c
(iv)  2sin 9A sin4A
(vii) 2 cos50° sin30°
(vi) 2 cos37° cos 14°
(ix)2sin30°cosl0°
53° 17°
(x) 2 cos y cos —j
(1) (i)"
(ll) T (ill) T
EXERCISE 6.7
(iv)* (v)
(vi)
n n
7. (vn) 7
262
,„. ,.. W7T , ,,„ It ,.. N 71 ,... N 2n7t 7t
(2) (i) y + ( 1)" 12 W "" " 3 ( m ) "3" ± 4
71 ?27E 71
(3) (i) tin ± 7 , Wi (ii) y , (2n +1)2 (iii) "Tt
71 271 71 ?17l
(4) (i) 2htt ± T (ii) 2«7t + 71 (iii) 2«7t + y , (In + 1) T (iv) y , 2«7t
(5) (i) 2mtt + 4 (or) nn + ( l) n ^  4
7t 7C 271
(ii) 2«7t  v (iii) 2«7t  t (iv) 2«7t, 2rai + y
EXERCISE 6.9
u u n 3tt k 3k
(1) (1)3 (11)3 (111) 2 (iv)y (v)j (vi)y
(3) (i){§ (ii) (iii)y (iv)y (7)* = ±n
(8) 2V3 (9)* = ^
263
Objective Type Questions  Answers (Key)
(1) 1
(2)4
(3)2
(4)1
(5)2
(6)4
(7)1
(8)3
(9)1
(10)1
(11)1
(12)2
(13)1
(14)4
(15)1
(16)4
(17)2
(18)1
(19)3
(20)3
(21)2
(22)1
(23)4
(24)3
(25)1
(26)3
(27)2
(28)2
(29)2
(30)3
(31)1
(32)4
(33)4
(34)1
(35)1
(36)3
(37)3
(38)2
(39)2
(40)4
(41)4
(42)1
(43)1
(44)3
(45)1
(46)2
(47)1
(48)3
(49)1
(50)1
(51)3
(52)4
(53)1
(54)3
(55)3
(56)3
(57)1
(58)2
(59)3
(60)1
(61)2
(62)3
(63)2
(64)3
(65)3
(66)2
(67)1
(68)1
(69)4
(70)1
(71)3
(72)1
(73)3
(74)3
(75)3
(76)2
(77)3
(78)3
(79)4
(80)3
(81)1
(82)3
(83)4
(84)3
(85)2
(86)4
(87)3
(88)3
(89)4
(90)2
(91)3
(92)1
(93)3
(94)3
(95)3
(96)1
(97)2
(98)3
(99)4
(100) 2
(101)4
(102) 1
(103) 3
(104) 4
(105) 1
(106) 3
(107) 2
(108)2
(109) 2
(110)3
(111)1
264